README.md

283. 移动零

English Version

题目描述

给定一个数组 nums，编写一个函数将所有 0 移动到数组的末尾，同时保持非零元素的相对顺序。

示例:

输入: [0,1,0,3,12]
输出: [1,3,12,0,0]

说明:

1. 必须在原数组上操作，不能拷贝额外的数组。
2. 尽量减少操作次数。

解法

Python3

class Solution:
    def moveZeroes(self, nums: List[int]) -> None:
        """
        Do not return anything, modify nums in-place instead.
        """
        left, n = 0, len(nums)
        for right in range(n):
            if nums[right] != 0:
                nums[left], nums[right] = nums[right], nums[left]
                left += 1

Java

class Solution {
    public void moveZeroes(int[] nums) {
        int left = 0, n = nums.length;
        for (int right = 0; right < n; ++right) {
            if (nums[right] != 0) {
                int t = nums[left];
                nums[left] = nums[right];
                nums[right] = t;
                ++left;
            }
        }
    }
}

C++

class Solution {
public:
    void moveZeroes(vector<int>& nums) {
        int left = 0, n = nums.size();
        for (int right = 0; right < n; ++right) {
            if (nums[right] != 0)
            {
                swap(nums[left], nums[right]);
                ++left;
            }
        }
    }
};

Go

func moveZeroes(nums []int) {
	n := len(nums)
	left := 0
	for right := 0; right < n; right++ {
		if nums[right] != 0 {
			nums[left], nums[right] = nums[right], nums[left]
			left++
		}
	}
}

JavaScript

/**
 * @param {number[]} nums
 * @return {void} Do not return anything, modify nums in-place instead.
 */
var moveZeroes = function (nums) {
    let left = 0,
        n = nums.length;
    for (let right = 0; right < n; ++right) {
        if (nums[right]) {
            [nums[left], nums[right]] = [nums[right], nums[left]];
            ++left;
        }
    }
};

Rust

impl Solution {
    pub fn move_zeroes(nums: &mut Vec<i32>) {
        let mut i = 0;
        for j in 0..nums.len() {
            if nums[j] != 0 {
                if i != j {
                    nums[i] = nums[j];
                    nums[j] = 0;
                }
                i += 1;
            }
        }
    }
}

...