LeetCode #942 DI StringMatch. Algorithm,알고리즘,LeetCode,Codefights,CodeSignal,코드파이트,코드시그널,예제,문제해결능력,example,c++,java,재귀,recursive,datastructure,techinterview,coding,코딩인터뷰,기술면접,연결리스트
5min
Runtime: 40 ms, faster than 68.48% of C++ online submissions for DI String Match.
Memory Usage: 10.5 MB, less than 7.14% of C++ online submissions for DI String Match.
LeetCode #942
Q.
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]
string S 가 주어지는데, 'I' 아니면 'D' 로 이루어져 있다. S 의 length 는 N 이라한다.
0 ~ N 까지 숫자가 들어가는 어떤 순열을 반환해라
- If S[i] == "I", then A[i] < A[i+1] S[i] 번쨰 값이 'I' 이면 뒤의 수보다 작다.
- If S[i] == "D", then A[i] > A[i+1] S[i] 번째 값이 'D' 이면 앞의 수보다 크다.
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".
Process
// Process
//1. Input string S
//2. Check start int I, D
//3. Iterate from begin to the end of S (i)
// 3.1. Check if it's I or D
// 3.1.1. If it's I -> push_back(intI) and ++
// 3.1.1. If it's D -> push_back(intD) and --
//4. Return resultVector
// 처리과정
//1. 스트링 S 를 입력받는다.
//2. 시작 int I 와 D 를 정해둔다.
//3. S 의 시작부터 끝까지 반복한다.
// 3.1. I 인지 D 인지 확인해서
// 3.1.1. I 이면 -> intI 를 리턴벡터에 넣고 ++ 한다.
// 3.1.2. D 이면 -> intD 를 리턴벡터에 넣고 -- 한다.
//4. Return resultVector
Code.. lemme see example code!!!
코드.. 예제코드를 보자!!!
class Solution {
public:
vector<int> diStringMatch(string S) {
vector<int> resultVec;
int intI = 0;
int intD = S.length();
for (int i = 0; i < S.length(); ++i) {
if (S[i] == 'I') {
resultVec.push_back(intI++);
} else { // 'D'
resultVec.push_back(intD--);
}
}
resultVec.push_back(intI);
return resultVec;
}
};
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