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算法笔记: 力扣#283 移动零

admin 11月 12, 2020 0

问题描述

解法

分析

Python 实现

class :
    def moveZeroes(self, nums):
        """
        :type nums: List[int]
        :rtype: void Do not return anything, modify nums in-place instead.
        """
        point = 0
        for i in range(len(nums)):
            if nums[point]:
                point += 1
            elif nums[i]:
                nums[point], nums[i] = nums[i], 0
                point += 1

Java 实现

class  {
    public void moveZeroes(int[] nums) {
        int point = 0;
        for(int i = 0; i < nums.length; i++){
            if(nums[point] != 0)
                point ++;
            else if(nums[i] != 0){
                nums[point] = nums[i];
                nums[i] = 0;
                point ++;
            }
        }
    }
}

时间复杂度

O(N).

空间复杂度

O(1).

链接

283. Move Zeroes
283. 移动零
 (English version) Algorithm Notes: Leetcode#283 Move Zeroes