solution.md

# 串联所有单词的子串

<p>给定一个字符串&nbsp;<strong>s&nbsp;</strong>和一些长度相同的单词&nbsp;<strong>words。</strong>找出 <strong>s
    </strong>中恰好可以由&nbsp;<strong>words </strong>中所有单词串联形成的子串的起始位置。</p>
<p>注意子串要与&nbsp;<strong>words </strong>中的单词完全匹配，中间不能有其他字符，但不需要考虑&nbsp;<strong>words&nbsp;</strong>中单词串联的顺序。</p>
<p>&nbsp;</p>
<p><strong>示例 1：</strong></p>
<pre><strong>输入：  s =</strong> &quot;barfoothefoobarman&quot;,<strong>  words = </strong>[&quot;foo&quot;,&quot;bar&quot;]<strong><br />输出：</strong>[0,9]<strong><br />解释：</strong>从索引 0 和 9 开始的子串分别是 &quot;barfoo&quot; 和 &quot;foobar&quot; 。输出的顺序不重要, [9,0] 也是有效答案。</pre>
<p><strong>示例 2：</strong></p>
<pre><strong>输入：  s =</strong> &quot;wordgoodgoodgoodbestword&quot;,<strong>  words = </strong>[&quot;word&quot;,&quot;good&quot;,&quot;best&quot;,&quot;word&quot;]<strong><br />输出：</strong>[]</pre>

以下程序实现了这一功能，请你填补空白处内容：

```java
class Solution {
	public List<Integer> findSubstring(String s, String[] words) {
		List<Integer> res = new ArrayList<>();
		if (s == null || s.length() == 0 || words == null || words.length == 0)
			return res;
		HashMap<String, Integer> map = new HashMap<>();
		int one_word = words[0].length();
		int word_num = words.length;
		int all_len = one_word * word_num;
		for (String word : words) {
			map.put(word, map.getOrDefault(word, 0) + 1);
		}
		for (int i = 0; i < one_word; i++) {
			int left = i, right = i, count = 0;
			HashMap<String, Integer> tmp_map = new HashMap<>();
			while (right + one_word <= s.length()) {
				String w = s.substring(right, right + one_word);
				right += one_word;
				if (!map.containsKey(w)) {
					count = 0;
					left = right;
					tmp_map.clear();
				} else {
					tmp_map.put(w, tmp_map.getOrDefault(w, 0) + 1);
					count++;
					while (tmp_map.getOrDefault(w, 0) > map.getOrDefault(w, 0)) {
						______________________;
					}
					if (count == word_num)
						res.add(left);
				}
			}
		}
		return res;
	}
}
```

## template

```java
class Solution {
	public List<Integer> findSubstring(String s, String[] words) {
		List<Integer> res = new ArrayList<>();
		if (s == null || s.length() == 0 || words == null || words.length == 0)
			return res;
		HashMap<String, Integer> map = new HashMap<>();
		int one_word = words[0].length();
		int word_num = words.length;
		int all_len = one_word * word_num;
		for (String word : words) {
			map.put(word, map.getOrDefault(word, 0) + 1);
		}
		for (int i = 0; i < one_word; i++) {
			int left = i, right = i, count = 0;
			HashMap<String, Integer> tmp_map = new HashMap<>();
			while (right + one_word <= s.length()) {
				String w = s.substring(right, right + one_word);
				right += one_word;
				if (!map.containsKey(w)) {
					count = 0;
					left = right;
					tmp_map.clear();
				} else {
					tmp_map.put(w, tmp_map.getOrDefault(w, 0) + 1);
					count++;
					while (tmp_map.getOrDefault(w, 0) > map.getOrDefault(w, 0)) {
						String t_w = s.substring(left, left + one_word);
						count--;
						tmp_map.put(t_w, tmp_map.getOrDefault(t_w, 0) - 1);
						left += one_word;
					}
					if (count == word_num)
						res.add(left);
				}
			}
		}
		return res;
	}
}
```

## 答案

```java
String t_w = s.substring(left, left + one_word);
count--;
tmp_map.put(t_w, tmp_map.getOrDefault(t_w, 0) - 1);
left += one_word;
```

## 选项

### A

```java
String t_w = s.substring(left, left + one_word);
count--;
tmp_map.put(t_w, tmp_map.getOrDefault(t_w, 0));
left += one_word;
```

### B

```java
String t_w = s.substring(left, left + one_word);
count--;
tmp_map.put(t_w, tmp_map.getOrDefault(t_w, 0) + 1);
left += one_word;
```

### C

```java
String t_w = s.substring(left, left - one_word);
count--;
tmp_map.put(t_w, tmp_map.getOrDefault(t_w, 0) + 1);
left += one_word;
```