Leetcode: 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.

For example, given nums = [0, 1, 0, 3, 12], after calling your function, nums should be [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.

第二遍做法： 双指针压缩法：O(N), 空间：O(1)

实际上就是将所有的非0数向前尽可能的压缩，最后把没压缩的那部分全置0就行了。比如103040，先压缩成134，剩余的3为全置为0。过程中需要一个指针记录压缩到的位置。

public class Solution {
    public void moveZeroes(int[] nums) {
        if (nums==null || nums.length==0) return;
        int cur=0;
        for (int i=0; i<nums.length; i++) {
            if (nums[i]!=0) {
                nums[cur++] = nums[i];
            }
        }
        for (int i=cur; i<nums.length; i++) {
            nums[i] = 0;
        }
    }
}

Q：如果要把所有的0放在前面而不是后面呢？
A：同样的解题思路，但是是从后向前遍历，将非0数字压缩到后面。

双指针同向法：O(N^2),不是太好

Two pointer l,r, 

l is to find each 0 in the list, r is to start from l, find first non-0

l, r should not both start from 0, because if r is to the left of l, we don't want to swap it. example: [1, 0]

public class Solution {
    public void moveZeroes(int[] nums) {
        if (nums==null && nums.length==0) return;
        int l=0, r=0;
        while (l < nums.length && r < nums.length) {
            while (l<nums.length && nums[l]!=0) {
                l++;
            }
            r = l;
            while (r<nums.length && nums[r]==0) {
                r++;
            }
            if (l == nums.length || r == nums.length) break;
            swap(nums, l, r);
        }
    }
    
    public void swap(int[] nums, int l, int r) {
        int temp = nums[l];
        nums[l] = nums[r];
        nums[r] = temp;
    }
}