LeetCode周赛#111 Q3 DI String Match

题目来源：https://leetcode.com/contest/weekly-contest-111/problems/di-string-match/

问题描述

 942. DI String Match

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".

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题意

规定一个长度为N，由”I”和”D”组成的字符数组S，对应一个由0~N组成的N+1位整数数组A，S[i] = ”I”代表A[i] < A[i+1], S[i] = “D”代表A[i] > A[i+1]. 给定S，求任意一个符合规定的A.

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思路

S[i] = 第ni个‘I’对应A[i] = ni, S[i] = 第nd个‘D’对应A[i] = N-nd.

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代码

class Solution {
public:
    vector<int> diStringMatch(string S) {
        int n = S.size(), i = 0, cnti = 0, cntd = 0;
        int * v = new int[n+1];
        for (i=0; i<n; i++)
        {
            if (S[i] == 'I')
            {
                v[i] = cnti;
                cnti++;
            }
            else
            {
                v[i] = n - cntd;
                cntd++;
            }
        }
        v[n] = cnti;
        vector<int> V(v, v+n+1);
        delete[] v;
        return V;
    }
};