Facebook Phone Interview -- Move Zeros to Right (Easy)

最新推荐文章于 2025-06-04 08:28:43 发布

原创最新推荐文章于 2025-06-04 08:28:43 发布 · 341 阅读

0 ·

CC 4.0 BY-SA版权

LeetCode 题解专栏收录该内容

288 篇文章

订阅专栏

本文介绍了一种将数组中所有零移动到末尾同时保持非零元素原有顺序的算法实现。该算法采用双指针技术，通过一次遍历完成元素的重新排列，并提供了完整的C++代码示例及测试。

March 22th, 2016.

Move Zeros to the right end but keep the order of non-zero numbers.

#include <stdio.h>
#inlcude <iostream>
#include <vector>
using namespace std;

// move Zeros to the right end of the array.
// classical two pointers problem.
void moveZeroToRight(vector<int>& array) {
    if(array.size() <= 1) return; // if the array only has 0-1 values, no need to sort.
    int left = 0; // one pointer to remember the left copy.
    int i = 0; // one pointer to loop the array.
    while(i < array.size()) {
        if(array[i] != 0) {
            array[left++] = array[i++]; // if it is not zero, copy it to the left end.
        } else {
            i++;  // otherwise, i pointer keeps on moving.
        }
    }
    // copy all non-zeros to the left now. But we need to remember to set the left values to 0.
    while(left < array.size()) {
         array[left++] = 0;
    }
}

// the following is for test purpose.
void printArray(vector<int> array) {
    for(int i = 0; i < array.size(); ++i) {
         printf("%d\n", array[i]);
    }
}

int main(void) {
     vector<int> array;
     array.push_back(0); array.push_back(1);
     moveZeroToRight(array);
     printArray(array);
}