#Leetcode# 942. DI String Match

佚名 6年前 (2019-04-12) 随笔 1954人围观抢沙发百度已收录

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

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

代码：

class Solution {
public:
    vector<int> diStringMatch(string S) {
        vector<int> ans;
        int n = S.length();
        int numi, numd;
        for(int i = 0; i < n; i ++) {
            if(S[i] == 'I') numi ++;
            else numd ++;
        }
        if(numi == n) {
            for(int i = 0; i <= n; i ++)
                ans.push_back(i);
        } else if(numd == n) {
            for(int i = n; i >= 0; i --)
                ans.push_back(i);
        } else {
            int st = 0, en = n;
            for(int i = 0; i < n; i ++) {
                if(S[i] == 'I') {
                    ans.push_back(st);
                    st ++;
                } else {
                    ans.push_back(en);
                    en --;
                }
            }
            ans.push_back(st);
        }
        return ans;
    }
};

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