942. DI String Match - easy
문제
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]
제한사항
- 1 <= S.length <= 10000
- S only contains characters “I” or “D”.
입출력 예
1
2
3
4
5
6
7
8
9
10
11
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]
풀이
- Math
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
class Solution {
public:
vector<int> diStringMatch(string S) {
vector<int> item(S.size() + 1, 0);
int low = 0, high = S.size();
for(auto i = 0 ; i < S.size() ; ++i){
if(S[i] == 'I'){
item[i] = low++;
}
else{
item[i] = high--;
}
}
item[S.size()] = low;
return item;
}
};