942

942. DI String Match (Easy)

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".

Solution 1.

// OJ: https://leetcode.com/problems/di-string-match/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
    vector<int> diStringMatch(string S) {
        vector<int> ans(S.size() + 1, 0);
        int lo = 0, hi = S.size();
        for (int i = 0; i < S.size(); ++i) {
            ans[i] = S[i] == 'I' ? lo++ : hi--;
        }
        ans[S.size()] = lo;
        return ans;
    }
};