README_EN.md

comments	difficulty	edit_url	tags
true	Easy	https://github.com/doocs/leetcode/edit/main/solution/0200-0299/0283.Move%20Zeroes/README_EN.md	Array Two Pointers

283. Move Zeroes

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Description

Given an integer array nums, move all 0's to the end of it while maintaining the relative order of the non-zero elements.

Note that you must do this in-place without making a copy of the array.

 

Example 1:

Input: nums = [0,1,0,3,12]
Output: [1,3,12,0,0]

Example 2:

Input: nums = [0]
Output: [0]

 

Constraints:

• 1 <= nums.length <= 104
• -231 <= nums[i] <= 231 - 1

 

Follow up: Could you minimize the total number of operations done?

Solutions

Solution 1: Two Pointers

We use a pointer $k$ to record the current position to insert, initially $k = 0$.

Then we iterate through the array $\textit{nums}$, and each time we encounter a non-zero number, we swap it with $\textit{nums}[k]$ and increment $k$ by 1.

This way, we can ensure that the first $k$ elements of $\textit{nums}$ are non-zero, and their relative order is the same as in the original array.

The time complexity is $O(n)$, where $n$ is the length of the array $\textit{nums}$. The space complexity is $O(1)$.

Python3

class Solution:
    def moveZeroes(self, nums: List[int]) -> None:
        k = 0
        for i, x in enumerate(nums):
            if x:
                nums[k], nums[i] = nums[i], nums[k]
                k += 1

Java

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

C++

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

Go

func moveZeroes(nums []int) {
	k := 0
	for i, x := range nums {
		if x != 0 {
			nums[i], nums[k] = nums[k], nums[i]
			k++
		}
	}
}

TypeScript

/**
 Do not return anything, modify nums in-place instead.
 */
function moveZeroes(nums: number[]): void {
    let k = 0;
    for (let i = 0; i < nums.length; ++i) {
        if (nums[i]) {
            [nums[i], nums[k]] = [nums[k], nums[i]];
            ++k;
        }
    }
}

Rust

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

JavaScript

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

C

void moveZeroes(int* nums, int numsSize) {
    int k = 0;
    for (int i = 0; i < numsSize; ++i) {
        if (nums[i] != 0) {
            int t = nums[i];
            nums[i] = nums[k];
            nums[k++] = t;
        }
    }
}