LeetCode #1089 DuplicateZeros. Algorithm,알고리즘,LeetCode,Codefights,CodeSignal,코드파이트,코드시그널,예제,그래프,Graph,example,c++,java,재귀,recursive,datastructure,techinterview,coding,코딩인터뷰,기술면접
Runtime: 36 ms, faster than 30.41% of C++ online submissions for Duplicate Zeros.
Memory Usage: 9.6 MB, less than 100.00% of C++ online submissions for Duplicate Zeros.
LeetCode #1089
Q.
Given a fixed length array arr of integers, duplicate each occurrence of zero, shifting the remaining elements to the right.
Note that elements beyond the length of the original array are not written.
Do the above modifications to the input array in place, do not return anything from your function.
고정된 길이의 주어진 정수 배열 arr 에서, 0이 있으면 0을 복사해라, 남은 정수들은 오른쪽으로 이동한다.
기존 길이를 넘어서는 것들은 사라진다.
반환값 없이 주어진 배열을 수정해라.
Example 1:
Input: [1,0,2,3,0,4,5,0]
Output: null
Explanation: After calling your function, the input array is modified to: [1,0,0,2,3,0,0,4]
Example 2:
Input: [1,2,3]
Output: null
Explanation: After calling your function, the input array is modified to: [1,2,3]
Note:
- 1 <= arr.length <= 10000
- 0 <= arr[i] <= 9
Process
// Process
//1. Input integer vector
//2. Iterate all
// 2.1. Check if it's 0 or not
// 2.1.1. If so -> add 0, erase end of vector
//3. Finish
// 처리과정
//1. 정수 배열 입력받는다.
//2. 전체 반복한다.
// 2.1. 0인지 확인해서
// 2.1.1. 0이면 0을 한개 더 넣고, 배열 끝 값을 지운다.
//3. 끝낸다.
Code.. lemme see example code!!!
코드.. 예제코드를 보자!!!
class Solution {
public:
void duplicateZeros(vector<int>& arr) {
for (int i = 0; i < arr.size(); ++i) {
if (arr[i] == 0) {
arr.insert(arr.begin()+i, 0);
arr.erase(arr.begin()+arr.size()-1);
++i;
}
}
}
};
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