Leetcode 283. Move Zeroes java解法

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:

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

Approach #1 Bubble sort，我的解法,慢

class Solution {
    public void moveZeroes(int[] nums) {
        for(int i=0;i<nums.length;i++){
            for(int j=i+1;j<nums.length;j++){
                if(nums[i]==0&&j<nums.length){
                    int temp=nums[i];
                    nums[i]=nums[j];
                    nums[j]=temp;
                }
            }
        }   
    }
}

时间复杂度O(N^2)

空间复杂度O(1)

改进：

public void moveZeroes(int[] nums) {

    int j = 0;
    for(int i = 0; i < nums.length; i++) {
        if(nums[i] != 0) {
            int temp = nums[j];
            nums[j] = nums[i];
            nums[i] = temp;
            j++;
        }
    }
}
}

Approach #2 Insert Index

// Shift non-zero values as far forward as possible
// Fill remaining space with zeros

public void moveZeroes(int[] nums) {
    if (nums == null || nums.length == 0) return;        

    int insertPos = 0;
    for (int num: nums) {
        if (num != 0) nums[insertPos++] = num;
    }        

    while (insertPos < nums.length) {
        nums[insertPos++] = 0;
    }
}

先把非零元素放到应该在的地方，再根据array长度算出结尾需要补的零，补在后面即可。

时间复杂度O(N)

空间复杂度O(1)

Approach #3 Snowball

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

像滚雪球一样，把0 都滚到一起，然后最左侧的零和非零元素对调。

时间复杂度O(N)

空间复杂度O(1)