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

思路：遇到D就拿前面的，遇到I就拿後面的；先要判斷有多少連續的D或者I

class Solution:
    def diStringMatch(self, S):
        """
        :type S: str
        :rtype: List[int]
        """
#        from collections import deque
#        q=deque([i for i in range(len(S)+1)])
        a=[i for i in range(len(S)+1)]
        p,q=0,len(a)-1
        res=[]
        if S[0]=='I': 
            res.append(a[p])
            p+=1
        else: 
            res.append(a[q])
            q-=1
        
        i=0
        while i<len(S):
            j=i
            while j<len(S) and S[j]==S[i]: j+=1
            if S[i]=='I':
                for k in range(j-i):
                    res.append(q-(j-i-1)+k)
                q-=(j-i)
            else:
                for k in range(j-i):
                    res.append(p+(j-i-1-k))
                p+=(j-i)
            i=j
        
        return res