# 942.Di String Match

**942.Di String Match**

> 给定只含 "I"（增大）或 "D"（减小）的字符串 S ，令 N = S.length。 返回 \[0, 1, ..., N] 的任意排列 A 使得对于所有 i = 0, ..., N-1，都有： 如果 S\[i] == "I"，那么 A\[i] < A\[i+1] 如果 S\[i] == "D"，那么 A\[i] > A\[i+1]

```
示例 1：

输出："IDID"
输出：[0,4,1,3,2]
示例 2：

输出："III"
输出：[0,1,2,3]
示例 3：

输出："DDI"
输出：[3,2,0,1]

提示：
1 <= S.length <= 1000
S 只包含字符 "I" 或 "D"。
```

方法： D表示大于，初始时候从0到N，全为I，当中间出现D的时候，可以将最大的n个数放在n个D处，满足条件。第一个D出现的位置是最大值，第二个是次大值...以此类推，代码如下：

```
class Solution {
public:
    vector<int> diStringMatch(string S) {
        vector<int> key;
        int max=S.length(),min=0;
        for(int i=0;i<S.length()+1;i++)
        {
            if(S[i]=='D')
                key.push_back(max--);
            else
                key.push_back(min++);
           }

        return key;
    }
};
```


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