leetcode 942. 增减字符串匹配(DI String Match)

题目描述:

给定只含 "I"(增大)或 "D"(减小)的字符串 S ,令 N = S.length

返回 [0, 1, ..., N] 的任意排列 A 使得对于所有 i = 0, ..., N-1,都有:

如果 S[i] == "I",那么 A[i] < A[i+1]
如果 S[i] == "D",那么 A[i] > A[i+1]

示例 1:

输出:"IDID"
输出:[0,4,1,3,2]

示例 2:

输出:"III"
输出:[0,1,2,3]

示例 3:

输出:"DDI"
输出:[3,2,0,1]

提示:

  1. 1 <= S.length <= 1000
  2. S 只包含字符 "I""D"

解法:

class Solution {
public:
    vector<int> diStringMatch(string S) {
        //ID
        //0 2 1
        //1 2 0
        //II
        //0 1 2
        //DD
        //2 1 0
        //DID
        //3 0 2 1
        // greedy algorithm
        int l = 0, r = S.size();
        vector<int> res;
        for(char ch : S){
            if(ch == 'I'){
                res.push_back(l++);
            }else{
                res.push_back(r--);
            }
        }
        res.push_back(l);
        return res;
    }
};

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转载自www.cnblogs.com/zhanzq/p/10670321.html