> For the complete documentation index, see [llms.txt](https://dfine.gitbook.io/leetcode/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://dfine.gitbook.io/leetcode/821.shortest_distance_to_a_character.md).

# 821.Shortest Distance to a Character

**821.Shortest Distance to a Character**

难度:Easy

> 给定一个字符串 S 和一个字符 C。返回一个代表字符串 S 中每个字符到字符串 S 中的字符 C 的最短距离的数组。

示例 1:

```
输入: S = "loveleetcode", C = 'e'
输出: [3, 2, 1, 0, 1, 0, 0, 1, 2, 2, 1, 0]
说明:
```

字符串 S 的长度范围为 \[1, 10000]。 C 是一个单字符，且保证是字符串 S 里的字符。 S 和 C 中的所有字母均为小写字母。

代码如下：

```
class Solution {
public:
    vector<int> shortestToChar(string S, char C) {
        vector<int> cc;
        int len=S.size();
        vector<int> index;
        for(auto i:S)
            if(i==C) {cc.push_back(0);
                      index.push_back(cc.size()-1);}
            else cc.push_back(len);
        int start=1;
        for(int i=index[0]+1;i<len;i++)
            if(cc[i]) cc[i]=start++;
            else start=1;
        start=1;
        for(int i=index[index.size()-1]-1;i>=0;i--)
            if(cc[i]) cc[i]=min(cc[i],start++);
            else start=1;
        return cc;
    }
};
```


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