# 942. DI String Match

**<font color=red>难度: Easy</font>**

## 刷题内容

> 原题连接

* https://leetcode.com/problems/di-string-match/

> 内容描述

```
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]
 

Note:

1 <= S.length <= 10000
S only contains characters "I" or "D".
```

## 解题方案

> 思路 1
******- 时间复杂度: O(N)******- 空间复杂度: O(1)******


- 按照I出现的顺序依次assign 值 0 到 i
- 按照D出现的顺序依次assign 值 N 到 i+2
- 最后一个数字就是i+1

sb题没什么好说的

```python
class Solution:
    def diStringMatch(self, S):
        """
        :type S: str
        :rtype: List[int]
        """
        if not S or len(S) == 0:
            return [0]
        res = [0] * (len(S)+1)
        idx1 = 0
        idx2 = len(S)
        for i in range(len(S)):
            if S[i] == 'I':
                res[i] = idx1
                idx1 += 1
            else:
                res[i] = idx2
                idx2 -= 1
        res[-1] = idx1 
        return res
```
