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add leetcode solution # 30 #35 for java
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‎Common/README.md‎

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# Algorithm
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一些关于算法的总结
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让你从算法入门到放弃
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![](./Document/src/main/jpg/BB527E2D-79C4-4303-9039-CE78973812AF.jpg)
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## Leetcode专区
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LeetCode 网站题目解析 [详见](./LeetCode)
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## AHaAlgorithms 啊哈算法
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书籍内部经典案例的再现，使用java实现
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## NewCodes
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牛客网
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## Pintia
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拼题A
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## Document
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算法文章类 [详见wiki](https://github.com/HbnKing/Algorithm/wiki) 已开放编辑
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## 贡献
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如果你有更好代码或者是算法相关专题(当然我们也非常的重视文档),你可以fork 修改测试后 pull requset to me,如果你对这个玩意非常感兴趣，可以将你加入到
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collaborator.
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作为贡献者 你可以在这里留下你的名字.
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贡献请阅读([Contributing](./CONTRIBUTING.md))
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## 其他
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后期将加入一些机器学习的东西

‎Common/pom.xml‎

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<?xml version="1.0" encoding="UTF-8"?>
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<project xmlns="http://maven.apache.org/POM/4.0.0"
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xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
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xsi:schemaLocation="http://maven.apache.org/POM/4.0.0 http://maven.apache.org/xsd/maven-4.0.0.xsd">
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<parent>
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<artifactId>algorithm</artifactId>
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<groupId>com.hbn</groupId>
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<version>1.0.0</version>
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</parent>
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<modelVersion>4.0.0</modelVersion>
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<artifactId>common</artifactId>
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</project>

‎LeetCode/README.md‎

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|39|[Combination Sum](https://oj.leetcode.com/problems/combination-sum/)| [Java](src/main/java/indi/ours/algorithm/leetcode/algorithms/_1.java/combinationSum/combinationSum)|Medium|
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|38|[Count and Say](https://oj.leetcode.com/problems/count-and-say/)| [Java](./src/main/java/indi/ours/algorithm/leetcode/algorithms/_38.java)|Easy| |
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|37|[Sudoku Solver](https://oj.leetcode.com/problems/sudoku-solver/)| [Java](src/main/java/indi/ours/algorithm/leetcode/algorithms/_1.java/sudokuSolver/sudokuSolver)|Hard|
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|36|[Valid Sudoku](https://oj.leetcode.com/problems/valid-sudoku/)| [Java](src/main/java/indi/ours/algorithm/leetcode/algorithms/_1.java/validSudoku/validSudoku)|Easy|
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|35|[Search Insert Position](https://oj.leetcode.com/problems/search-insert-position/)| [Java](src/main/java/indi/ours/algorithm/leetcode/algorithms/_1.java/searchInsertPosition/searchInsertPosition), [Java](./algorithms/java/src/searchInsertPosition/searchInsertPosition.java)|Medium|
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|36|[Valid Sudoku](https://oj.leetcode.com/problems/valid-sudoku/)| [Java](./src/main/java/indi/ours/algorithm/leetcode/algorithms/_1.java/validSudoku/validSudoku)|Easy|
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|35|[Search Insert Position](https://oj.leetcode.com/problems/search-insert-position/)| [Java](./src/main/java/indi/ours/algorithm/leetcode/algorithms/_35.java) |Easy| Array Binary Search| O(N) | | |
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|34|[Search for a Range](https://oj.leetcode.com/problems/search-for-a-range/)| [Java](src/main/java/indi/ours/algorithm/leetcode/algorithms/_1.java/searchForRange/searchForRange), [Java](./algorithms/java/src/searchForRange/searchForRange.java)|Medium|
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|33|[Search in Rotated Sorted Array](https://oj.leetcode.com/problems/search-in-rotated-sorted-array/)| [Java](src/main/java/indi/ours/algorithm/leetcode/algorithms/_1.java/searchInRotatedSortedArray/searchInRotatedSortedArray), [Java](./algorithms/java/src/searchInRotatedSortedArray/searchInRotatedSortedArray.java)|Hard|
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|32|[Longest Valid Parentheses](https://oj.leetcode.com/problems/longest-valid-parentheses/)| [Java](src/main/java/indi/ours/algorithm/leetcode/algorithms/_1.java/longestValidParentheses/longestValidParentheses)|Hard| String Dynamic Programming |
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|31|[Next Permutation](https://oj.leetcode.com/problems/next-permutation/)| [Java](src/main/java/indi/ours/algorithm/leetcode/algorithms/_1.java/nextPermutation/nextPermutation)|Medium|
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|30|[Substring with Concatenation of All Words](https://oj.leetcode.com/problems/substring-with-concatenation-of-all-words/)| [Java](src/main/java/indi/ours/algorithm/leetcode/algorithms/_1.java/substringWithConcatenationOfAllWords/substringWithConcatenationOfAllWords)|Hard| String|
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|30|[Substring with Concatenation of All Words](https://oj.leetcode.com/problems/substring-with-concatenation-of-all-words/)| [Java](./src/main/java/indi/ours/algorithm/leetcode/algorithms/_30.java)|Hard| String Hash Table Two Pointer| O(M*N)| | |
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|29|[Divide Two Integers](https://oj.leetcode.com/problems/divide-two-integers/)| [Java](src/main/java/indi/ours/algorithm/leetcode/algorithms/_1.java/divideTwoInt/divideTwoInt)|Medium|
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|28|[Implement strStr()](https://oj.leetcode.com/problems/implement-strstr/)| [Java](src/main/java/indi/ours/algorithm/leetcode/algorithms/_1.java/strStr/strStr), [Scala](./algorithms/java/src/strStr/strStr.java)|Easy|
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|27|[Remove Element](https://oj.leetcode.com/problems/remove-element/)| [Java](src/main/java/indi/ours/algorithm/leetcode/algorithms/_1.java/removeElement/removeElement)|Easy|
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|22|[Generate Parentheses](https://oj.leetcode.com/problems/generate-parentheses/)| [Java](src/main/java/indi/ours/algorithm/leetcode/algorithms/_1.java/generateParentheses/generateParentheses)|Medium|
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|21|[Merge Two Sorted Lists](https://oj.leetcode.com/problems/merge-two-sorted-lists/)| [Java](src/main/java/indi/ours/algorithm/leetcode/algorithms/_1.java/mergeTwoSortedList/mergeTwoSortedList)|Easy|
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|20|[Valid Parentheses](https://oj.leetcode.com/problems/valid-parentheses/)| [Java](./src/main/java/indi/ours/algorithm/leetcode/algorithms/_20.java)|Easy|String Stack|O(n) | |
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|19|[Remove Nth Node From End of List](https://oj.leetcode.com/problems/remove-nth-node-from-end-of-list/)| [Java](./src/main/java/indi/ours/algorithm/leetcode/algorithms/_19.java)|Easy| Linked List Two Pointer| O(N) ||&gamma;|
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|19|[Remove Nth Node From End of List](https://oj.leetcode.com/problems/remove-nth-node-from-end-of-list/)| [Java](./src/main/java/indi/ours/algorithm/leetcode/algorithms/_19.java)|Medium| Linked List Two Pointer| O(N) ||&gamma;|
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|18|[4Sum](https://oj.leetcode.com/problems/4sum/)| [Java](./src/main/java/indi/ours/algorithm/leetcode/algorithms/_18.java)|Medium|Array HashTable TwoPointer | O(N^4) || Q15|
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|17|[Letter Combinations of a Phone Number](https://oj.leetcode.com/problems/letter-combinations-of-a-phone-number/)| [Java](./src/main/java/indi/ours/algorithm/leetcode/algorithms/_17.java)|Medium|String BackTracking|O(n^3) | | queue|
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|16|[3Sum Closest](https://oj.leetcode.com/problems/3sum-closest/)| [Java](./src/main/java/indi/ours/algorithm/leetcode/algorithms/_16.java)|Medium|Array Two Point | O(n^2) | | Q15|

‎LeetCode/src/main/java/indi/ours/algorithm/leetcode/algorithms/_30.java‎

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package indi.ours.algorithm.leetcode.algorithms;
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import java.util.ArrayList;
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import java.util.HashMap;
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import java.util.List;
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import java.util.Map;
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/**
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* @author wangheng
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* @create 2018-11-11 下午2:30
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* @desc
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* You are given a string, s, and a list of words, words,
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* that are all of the same length.
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* Find all starting indices of substring(s) in s that is a concatenation of each word in words exactly once and without any intervening characters.
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*
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* Example 1:
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*
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* Input:
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* s = "barfoothefoobarman",
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* words = ["foo","bar"]
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* Output: [0,9]
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* Explanation: Substrings starting at index 0 and 9 are "barfoor" and "foobar" respectively.
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* The output order does not matter, returning [9,0] is fine too.
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* Example 2:
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*
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* Input:
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* s = "wordgoodstudentgoodword",
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* words = ["word","student"]
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* Output: []
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*
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*
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**/
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public class _30 {
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/**
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*
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* @param s
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* @param words
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* @return
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*/
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public List<Integer> findSubstring(String s, String[] words) {
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int length = words[0].length();
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List<Integer> res = new ArrayList<>();
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if (s == null || words == null || words.length == 0 || s.length() < words[0].length() * words.length) return res;
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Map<String, Integer> map = new HashMap();
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Map<String, Integer> cur = new HashMap();
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//将 数据放入到 map 中
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for (String word: words) {
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if (!map.containsKey(word)) map.put(word, 0);
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map.put(word, map.get(word) + 1);
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}
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int m = words[0].length(), n = s.length(), num = words.length;
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for (int i = 0; i < m; i++) {
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int left = i, right = i, cnt = 0;
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cur = new HashMap();
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for (; right + m <= n; right += m) {
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String sub = s.substring(right, right + m);
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if (!map.containsKey(sub)){
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cur = new HashMap();
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cnt = 0;
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left = right + m;
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}
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else {
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if (!cur.containsKey(sub)) {
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cur.put(sub, 0);
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}
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cur.put(sub, cur.get(sub) + 1);
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cnt++;
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while (cur.get(sub) > map.get(sub)) {
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String leftSub = s.substring(left, left + m);
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cur.put(leftSub, cur.get(leftSub) - 1);
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cnt--;
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left += m;
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}
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if (cnt == num) res.add(left);
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}
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}
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}
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return res;
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}
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/**
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* I think the following code is self-explanatory enough.
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* We use an unordered_map<string, int> counts to record
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* the expected times of each word and another unordered_map<string, int> seen to record the times we have seen.
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* Then we check for every possible position of i.
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* Once we meet an unexpected word or the times of some word is larger than its expected times, we stop the check.
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* If we finish the check successfully, push i to the result indexes.
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*
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* Just build a map for the words and their relative count in L. Then we traverse through S to check whether there is a match.
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* @param s
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* @param words
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* @return
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*/
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public List<Integer> findSubstring2(String s, String[] words) {
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final Map<String, Integer> counts = new HashMap<>();
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for (final String word : words) {
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counts.put(word, counts.getOrDefault(word, 0) + 1);
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}
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final List<Integer> indexes = new ArrayList<>();
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final int n = s.length(), num = words.length, len = words[0].length();
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for (int i = 0; i < n - num * len + 1; i++) {
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final Map<String, Integer> seen = new HashMap<>();
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int j = 0;
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while (j < num) {
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final String word = s.substring(i + j * len, i + (j + 1) * len);
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if (counts.containsKey(word)) {
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seen.put(word, seen.getOrDefault(word, 0) + 1);
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if (seen.get(word) > counts.getOrDefault(word, 0)) {
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break;
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}
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} else {
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break;
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}
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j++;
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}
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if (j == num) {
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indexes.add(i);
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}
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}
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return indexes;
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}
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}

‎LeetCode/src/main/java/indi/ours/algorithm/leetcode/algorithms/_35.java‎

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package indi.ours.algorithm.leetcode.algorithms;
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/**
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* @author wangheng
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* @create 2018-11-11 下午11:31
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* @desc
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* Given a sorted array and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order.
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*
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* You may assume no duplicates in the array.
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*
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* Example 1:
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*
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* Input: [1,3,5,6], 5
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* Output: 2
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* Example 2:
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*
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* Input: [1,3,5,6], 2
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* Output: 1
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* Example 3:
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*
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* Input: [1,3,5,6], 7
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* Output: 4
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* Example 4:
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*
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* Input: [1,3,5,6], 0
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* Output: 0
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*
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* 找到目标应该 插入的位置
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*
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**/
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public class _35 {
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/**
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* O(N)
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* @param nums
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* @param target
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* @return
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*/
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public int searchInsert(int[] nums, int target) {
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for(int i = 0 ;i<nums.length ;i++){
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if(nums[i] >=target){
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return i ;
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}
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}
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//如果便利完成后，还没有返回 ，说明数组 的内容都比 target 小
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//应该放在最后面
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return nums.length ;
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}
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}

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