# 串联所有单词的子串
给定一个字符串 s 和一些长度相同的单词 words。找出 s
中恰好可以由 words 中所有单词串联形成的子串的起始位置。
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。
示例 1：
输入： s = "barfoothefoobarman", words = ["foo","bar"]
输出：[0,9]
解释：从索引 0 和 9 开始的子串分别是 "barfoo" 和 "foobar" 。输出的顺序不重要, [9,0] 也是有效答案。
示例 2：
输入： s = "wordgoodgoodgoodbestword", words = ["word","good","best","word"]
输出：[]
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 findSubstring(String s, String[] words) {
List res = new ArrayList<>();
if (s == null || s.length() == 0 || words == null || words.length == 0)
return res;
HashMap 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 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 findSubstring(String s, String[] words) {
List res = new ArrayList<>();
if (s == null || s.length() == 0 || words == null || words.length == 0)
return res;
HashMap 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 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;
```