LeetCode刷题笔记--942. DI String Match

942. DI String Match

Easy

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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".

解法：这个算今天做到最简单的题了。

class Solution {
public:
    vector<int> diStringMatch(string S) {
        vector<int> ans;
        int N=S.length();
        int hi=N;
        int lo=0;
        for(int j=0;j<=N;j++)
        {
            if(S[j]=='I')ans.push_back(lo++);
            else ans.push_back(hi--);
        }
        return ans;
    }
};