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".

Approach #1:

class Solution {
public:
    vector<int> diStringMatch(string S) {
        int len = S.length();
        vector<int> ans(len+1, 0);
        int index = 0;
        for (int i = 0; i < len; ++i)
            if (S[i] == 'I') ans[i] = index++;
        ans[len] = index++;
        for (int i = len-1; i >= 0; --i)
            if (S[i] == 'D') ans[i] = index++;
        return ans;     
    }
};