283. Move Zeroes

题目：

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:

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

链接： http://leetcode.com/problems/move-zeroes/

题解：

把0都移动到队尾。这道题在Bloomberg的电面里还真被问到过。

Time Complexity - O(n)， Space Complexity - O(1)。

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

二刷：

Java:

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

三刷:

先用一个int来count有多少个非0元素，然后把count ~ 数组末尾的所有元素置0.

Java:

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

Update:

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

Reference: