LeetCode-4.4-美区-283-E-移动零（Move Zeroes)

Given an array nums, write a function to move all 0’s to the end of it while maintaining the relative order of the non-zero elements.

Example:
Input: [0,1,0,3,12]
Output: [1,3,12,0,0]

Note:
You must do this in-place without making a copy of the array.
Minimize the total number of operations.

思路

（1）题目后半段读错了，以为题目的意思是把0移到后面，前面的元素按照升序排列:(，一定程度受到示例的影响。所以先让数组降序，然后再对前面非0部分做了首尾交换的升序排列。
（2）题目真正的意思：maintaining the relative order of the non-zero elements. 保持非零元素的相对位置不变。
（3）参见 遵守循环不变式（Java）——by liweiwei1419

解法1-错误

class Solution {
    public void moveZeroes(int[] nums) {
        int len = nums.length;
        sort(nums, 0, len-1);
        int i=0;
        for(; i < len; i++){
            if(nums[i]==0){
                break;
            }
        }
        i--;
        int j = 0;
        while(j < i){
            int tmp = nums[i];
            nums[i] =nums[j];
            nums[j] = tmp;
            i--;
            j++;
        }
    }
    
    private void sort(int[] arr, int left, int right){
        if(left < right){
            int mid = quickSort(arr, left, right);
            sort(arr, left, mid-1);
            sort(arr, mid+1, right);
        }
    }

    private int quickSort(int[] arr, int left, int right){
        int tmp = arr[left];
        int i = left;
        int j = right;

        while(true){
            while(i<=j && arr[i] >= tmp){
                i++;
            }
            while(i<=j && arr[j] <= tmp){
                j--;
            }
            if(i >= j){
                break;
            }

            int temp = arr[i];
            arr[i] = arr[j];
            arr[j] = temp;
        }
        arr[left] = arr[j];
        arr[j] = tmp;
        return j;
    }
}

解法2

（1）统计0的个数；
（2）有几个0，就把当前元素往前移动几个位置，相当于和0互换位置。

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