LeetCode 283. 移动零

283. Move Zeroes

Given an integer array?nums, move all?0's to the end of it while maintaining the relative order of the non-zero elements.

Note?that you must do this in-place without making a copy of the array.

Example 1:

Input: nums = [0,1,0,3,12]

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

Example 2:

Input: nums = [0]

Output: [0]

Constraints:

• 1 <= nums.length <= 10^4
• -2^31?<= nums[i] <= 2^31?- 1

Follow up:?Could you minimize the total number of operations done?

思考思路：

1. loop, count zeros （代码方法三）

2. 开新数组， loop（此方法比较戳）

3. index（代码方法一二）

方法一：

class Solution {
    public void moveZeroes(int[] nums) {
        // index
        // Time: O(n) Space: O(1)
        int j = 0;
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] != 0) {
                nums[j] = nums[i];
                if (i != j) {
                    nums[i] = 0;
                }
                j++;
            }           
        }
    }
}

方法二：

class Solution {
    public void moveZeroes(int[] nums) {
        // Time: O(n) Space: O(1)
        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;
            }
        }
    }
}

方法三：

class Solution {
    public void moveZeroes(int[] nums) {
        // Time: O(n) Space: O(1)
        if (nums == null || nums.length == 0) {
            return;
        }
        // 第一次遍历讲非0元素移动到最前面，j指针记录非零的个数
        int j = 0;
        for (int i = 0; i < nums.length; ++i) {
            if (nums[i] != 0) {
                nums[j++] = nums[i];
            }
        }
        // 第二次遍历将非0后面的元素置零
        for (int i = j; i < nums.length; ++i) {
            nums[i] = 0;
        }
    }
}