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Commit baf4776

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Merge pull request #62 from Mrzhudky/work
Add Solution and Readme.md for 30[Java]
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‎solution/029.Divide Two Integers/README.md‎

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返回被除数 dividend 除以除数 divisor 得到的商。
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```
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示例 1:
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输入: dividend = 10, divisor = 3
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输入: dividend = 7, divisor = -3
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输出: -2
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```
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说明:
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## 与所有单词相关联的字串
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### 题目描述
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给定一个字符串 s 和一些长度相同的单词 words。在 s 中找出可以恰好串联 words 中所有单词的子串的起始位置。
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注意子串要与 words 中的单词完全匹配,中间不能有其他字符,但不需要考虑 words 中单词串联的顺序。
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```
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示例 1:
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输入:
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s = "barfoothefoobarman",
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words = ["foo","bar"]
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输出: [0,9]
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解释: 从索引 0 和 9 开始的子串分别是 "barfoor" 和 "foobar" 。
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输出的顺序不重要, [9,0] 也是有效答案。
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示例 2:
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输入:
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s = "wordgoodstudentgoodword",
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words = ["word","good","good"]
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(ps:原题的例子为 words = ["word","student"] 和题目描述不符,这里私自改了一下)
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输出: []
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```
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### 解法
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1. 用 HashMap< 单词, 出现次数 > map 来存储所有单词;
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2. 设单词数量为 N ,每个单词长度为 len,则我们只需要对比到 **str.length() - N \* len** ,
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再往后因为不足 N \* len 个字母,肯定不匹配;
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3. 每次从 str 中选取连续的 N \* len 个字母进行匹配时,**从后向前匹配**,因为若后面的单词不匹配,
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无论前面的单词是否匹配,当前选取的字串一定不匹配,且,最后一个匹配的单词前的部分一定不在匹配的字串中,
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这样下一次选取长度为 N \* len 的字串时,可以**从上次匹配比较中最后一个匹配的单词开始**,减少了比较的次数;
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4. 考虑到要点 3 中对前一次匹配结果的利用,遍历 str 时,采用间隔为 len 的形式。
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例如示例 1 ,遍历顺序为:(0 3 6 9 12 15) (1 4 7 10 13)(2 5 8 11 14)
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```java
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class Solution {
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public List<Integer> findSubstring(String s, String[] words) {
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List<Integer> re = new ArrayList<>();
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if(s == null || words == null || words.length == 0 || words[0] == null) {
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return re;
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}
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if(s.length() == 0 || words[0].length() == 0 || s.length() < words.length * words[0].length()) {
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return re;
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}
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// 用< 单词,出现次数 > 来存储 words 中的元素,方便查找
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HashMap<String,Integer> map = new HashMap();
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for (String string : words) {
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map.put(string, map.getOrDefault(string,0) + 1);
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}
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int len = words[0].length();
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int strLen = s.length();
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int lastStart = len * words.length - len;
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for (int i = 0; i < len; i++) {
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for (int j = i; j <= strLen - len - lastStart; j += len) {
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String tempStr = s.substring(j, j + len);
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if(map.containsKey(tempStr)) {
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HashMap<String,Integer> searched = new HashMap<>();
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// 从后向前依次对比
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int tempIndex = j + lastStart;
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String matchedStr = s.substring(tempIndex, tempIndex + len);
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while (tempIndex >= j && map.containsKey(matchedStr)) {
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// 正确匹配到单词
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if(searched.getOrDefault(matchedStr,0) < map.get(matchedStr)) {
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searched.put(matchedStr, searched.getOrDefault(matchedStr,0) + 1);
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}
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else {
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break;
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}
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tempIndex -= len;
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if(tempIndex < j) {
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break;
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}
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matchedStr = s.substring(tempIndex, tempIndex + len);
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}
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// 完全匹配所以单词
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if(j > tempIndex) {
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re.add(j);
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}
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// 从tempIndex 到 tempIndex + len 这个单词不能正确匹配
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else {
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j = tempIndex;
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}
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}
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}
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}
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return re;
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}
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}
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```
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class Solution {
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public List<Integer> findSubstring(String s, String[] words) {
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List<Integer> re = new ArrayList<>();
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if(s == null || words == null || words.length == 0 || words[0] == null) {
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return re;
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}
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if(s.length() == 0 || words[0].length() == 0 || s.length() < words.length * words[0].length()) {
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return re;
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}
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// 用< 单词,出现次数 > 来存储 words 中的元素,方便查找
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HashMap<String,Integer> map = new HashMap();
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for (String string : words) {
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map.put(string, map.getOrDefault(string,0) + 1);
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}
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int len = words[0].length();
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int strLen = s.length();
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int lastStart = len * words.length - len;
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for (int i = 0; i < len; i++) {
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for (int j = i; j <= strLen - len - lastStart; j += len) {
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String tempStr = s.substring(j, j + len);
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if(map.containsKey(tempStr)) {
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HashMap<String,Integer> searched = new HashMap<>();
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// 从后向前依次对比
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int tempIndex = j + lastStart;
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String matchedStr = s.substring(tempIndex, tempIndex + len);
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while (tempIndex >= j && map.containsKey(matchedStr)) {
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// 正确匹配到单词
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if(searched.getOrDefault(matchedStr,0) < map.get(matchedStr)) {
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searched.put(matchedStr, searched.getOrDefault(matchedStr,0) + 1);
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}
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else {
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break;
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}
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tempIndex -= len;
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if(tempIndex < j) {
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break;
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}
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matchedStr = s.substring(tempIndex, tempIndex + len);
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}
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// 完全匹配所以单词
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if(j > tempIndex) {
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re.add(j);
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}
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// 从tempIndex 到 tempIndex + len 这个单词不能正确匹配
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else {
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j = tempIndex;
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}
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}
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}
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}
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return re;
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}
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}

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