LeetCode 283. Move Zeroes

问题：

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.

分析：

双指针。这一问题的一个中间步骤，应该是前m个数有序，中间是n个0，最后l个数无序。第一个指针指向第一个0，第二个指针指向l个无序数中的第一个。

当后个指针指向最后一个数的后一个位置时，所有步骤结束。

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

结果：

Run Code Status: Runtime Error
Run Code Result:
Your input

[0,1,0,3,12]

Your answer

Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: 5
    at Solution.moveZeroes(Solution.java:5)
    at __DriverSolution__.__helper__(__Driver__.java:8)
    at __Driver__.main(__Driver__.java:52)

分析：

二次：

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

结果：

Submission Result: Wrong Answer 
 
Input: [1]
Output: [0]

Expected: [1]

分析：

还是不能直接赋值0.

三次：

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