283. 移动零

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:

必须在原数组上操作，不能拷贝额外的数组。
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.
        """
        if not nums:
            return
        n = len(nums)
        zero_count = 0
        for i in range(n):
            if nums[i] == 0:
                zero_count += 1
            else:
                nums[i - zero_count] = nums[i]
        while zero_count > 0:
            nums[n - zero_count] = 0
            zero_count -= 1

Java

class Solution {
    public void moveZeroes(int[] nums) {
        int n;
        if (nums == null || (n = nums.length) < 1) {
            return;
        }
        int zeroCount = 0;
        for (int i = 0; i < n; ++i) {
            if (nums[i] == 0) {
                ++zeroCount;
            } else {
                nums[i - zeroCount] = nums[i];
            }
        }
        while (zeroCount > 0) {
            nums[n - zeroCount--] = 0;
        }
    }
}

JavaScript

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

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++
		}
	}
}

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