283. Move Zeroes - easy

自己的解法，用了两个pointer。速度比较慢。
答案很简单，用了一个pointer记录非零数insert的position, 并且用iteration做循环。

My code, 1st:

public class Solution {
    public void moveZeroes(int[] nums) {
        /** two pointers:
         * 1st first zero, 2nd curr scan iterm
         */
        int zero = -1, curr = 0;
        for (int i=0; i<nums.length; i++) {
            if (nums[i] != 0) continue;
            else {
                zero = i;
                break;
            }
        }
        if(zero == -1)  return;
        
        for (int i=zero+1; i<nums.length; i++) {
            if (nums[i]==0) continue;
            else {
                swap(nums, zero, i);
                zero ++;
            }
        }
    }
    
    private void swap(int[] nums, int a, int b) {
        int tmp = nums[a];
        nums[a] = nums[b];
        nums[b] = tmp;
    }
}

看了答案之后：
这种做法问题是改变了array本身。

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

想法：在scan的时候没有遇见0，就只向后移动。
does not matter, just 1ms.
时间: O(n);
空间：O(n);

Never stop. Oct 19 2016