(C++練習) 942. DI String Match

題目 :

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:

大意 :

針對 D/I 做一些題目規劃的運算, 則第一個遇到的D 會是總長度.

class Solution {
public:
    vector<int> diStringMatch(string S) {

        int len = strlen(S.c_str());
        char symbolD = 'D';
        char symbolI = 'I';
        int lo = 0, hi = len;
        vector<int> ans;
        
        for (int i = 0; i < len; i++){

            if (S[i] == symbolI)
                ans.push_back(lo++);
            else
                ans.push_back(hi--);
        }
        ans.push_back(lo);
        return ans;
    }
};