# 942. DI String Match

Given a string `S` that **only** contains "I" (increase) or "D" (decrease), let `N = S.length`.

Return **any** permutation `A` of `[0, 1, ..., N]` such that for all `i = 0, ..., N-1`:

* If `S[i] == "I"`, then `A[i] < A[i+1]`
* If `S[i] == "D"`, then `A[i] > A[i+1]`

**Example 1:**

```
Input: "IDID"
Output: [0,4,1,3,2]
```

**Example 2:**

```
Input: "III"
Output: [0,1,2,3]
```

**Example 3:**

```
Input: "DDI"
Output: [3,2,0,1]
```

#### 解题要点：

遍历字符串（转换成数组），遇D，从字符串长度开始加；遇I，从0开始加。最后加入从零开始的那个值，返回数组。

```java
class Solution {
    public int[] diStringMatch(String S) {
        int N = S.length();
        int[] res = new int[N + 1];
        if(N == 0) return res;
        int start = 0, count = 0;
        for(char c : S.toCharArray()){
            if(c == 'I') res[count++] = start++;
            if(c == 'D') res[count++] = N--;
        }
        res[count] = start;
        return res;
    }
}
```