9. DI String Match

Title：

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]

Note:

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

Analysis of Title：

If S[i] == "I", then A[i] < A[i+1]　　It means when get a 'I' then in here is a smaller and the next is bigger.

If S[i] == "D", then A[i] > A[i+1]　　In the similar way.

Test case：

"IDID"

Python：

ps: A simple way is "for : If I then get 0,0+1; If D then get len(S),len(S)-1"

But I want to learn another method of deque.

class Solution(object):
　　def diStringMatch(self, S):
　　"""
　　:type S: str
　　:rtype: List[int]
　　"""
　　from collections import deque

　　table = deque(range(len(S)+1)) #创建双端队列可迭代对象,数据包括 0~len(S)+1
　　res = []
　　for s in S:
　　　　if s=='I':
　　　　　　res.append(table.popleft()) #若为I，加入左边（小）
　　　　else:
　　　　　　res.append(table.pop()) #若为D，加入右边（大）
　　res.append(table.pop()) #把最后一个加进去，输出比输入多一位
　　return res

Analysis of Code：

All the analysis is already in the Title.