版权声明:本文为博主原创文章,未经博主允许不得转载。@ceezyyy11 https://blog.csdn.net/ceezyyy11/article/details/89218692
[LeetCode] 942. DI String Match (C++)
Easy
Share
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 <= S.length <= 10000
S only contains characters “I” or “D”.
class Solution {
public:
vector<int> diStringMatch(string S) {
vector<int> res;
int i=0; // min
int j=S.size(); // max
for(char c:S) {
if(c=='D') { // decrease
res.push_back(j--);
}
else { // increase
res.push_back(i++);
}
}
res.push_back(i); // the last element
return res;
}
};
/**
* Submission Detail:
* Runtime: 40 ms, faster than 99.76% of C++ online submissions for DI String Match.
* Memory Usage: 10.3 MB, less than 98.79% of C++ online submissions for DI String Match.
*/