LeetCode #283 MoveZeroes #2. Algorithm,알고리즘,LeetCode,Codefights,CodeSignal,코드파이트,코드시그널,예제,문제해결능력,example,c++,java,재귀,recursive,datastructure,techinterview,coding,코딩인터뷰,기술면접
쉬운거 한개만 빨리 푸려고 아무거나 풀었는데 풀고 저장하려고 보니 예전에 풀었던거..
이번에 그냥 푼게 퍼포먼스 더 잘나옴.. (32ms -> 20ms)
I just picked and solved easy one, but it was what i was solved in the past...
Anyway, it's better performance. (32ms -> 20ms)
LeetCode #283
Q.
Given an array nums, write a function to move all 0's to the end of it while maintaining the relative order of the non-zero elements.
숫자 배열이 주어지고, 모든 0 을 제일 뒤로 몰아넣어라.
e.g.
Example:
Input: [0,1,0,3,12]
Output: [1,3,12,0,0]
Note:
You must do this in-place without making a copy of the array.
Minimize the total number of operations.
추가로 :
배열 복사 없이 해야하고, 함수 사용을 최소화 해봐라.
Process
// Process
//1. input num vector
//2. Iterate from begin to the end
// 2.1. Check if it's 0
// 2.1.1. If so -> erase and count
//3. Iterate count number
// 3.1. Push_back zero to vector
//4. Finish
// 처리과정
//1. 배열을 입력받는다.
//2. 시작부터 끝까지 반복한다.
// 2.1. 0인지 확인해서
// 2.1.1. 0이면 -> 카운트센다. 지운다.
//3. 카운트 수만큼 반복한다.
// 3.1. 뒤에다 0을 넣어준다.
//4. 끝낸다.
Code.. lemme see example code!!!
코드.. 예제코드를 보자!!!
#include <vector>
using namespace std;
// Process
//1. input num vector
//2. Iterate from begin to the end - zeroCount
// 2.1. Check if it's 0
// 2.1.1. If so -> erase and push to back
//3. Return vector
class Solution {
public:
void moveZeroes(vector<int>& nums) {
int count = 0;
for (int i = 0; i < nums.size(); ++i)
{
if (nums[i] == 0)
{
++count;
nums.erase(nums.begin() + i);
--i;
}
}
for (int i = 0; i < count; ++i)
{
nums.push_back(0);
}
//int zeroCount = 0;
//for (int i = 0; i < nums.size() - zeroCount; ++i)
//{
// if (nums[i] == 0)
// {
// nums.erase(nums.begin() + i);
// nums.push_back(0);
// ++zeroCount;
// --i;
// }
//}
}
};
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