[LeetCode] #283 移动零

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。

输入: [0,1,0,3,12]

输出: [1,3,12,0,0]

类似于[LeetCode] #27 移除元素

先放非零数，再放零

class Solution {
    public void moveZeroes(int[] nums) {
        int index = 0,count = 0;
        for(int i = 0; i < nums.length; i++){
            if(nums[i] != 0) nums[index++] = nums[i];
            else count++;
        }
        for(int i = nums.length-1; i > nums.length-count-1; i--) nums[i]=0;
    }
}

双指针，零于非零交换

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

知识点：无

总结：无

标签：right,nums,int,++,length,283,移动,LeetCode,left
来源： https://www.cnblogs.com/jpppp/p/15292277.html