Mobile Zero of Likou Brush Questions

题目

给定一个数组 nums,编写一个函数将所有 0 移动到数组的末尾,同时保持非零元素的相对顺序.
请注意 ,必须在不复制数组的情况下原地对数组进行操作.

The question asks to move the zeros in the array to the back of the array.And can only operate on the original array,不可以复制,You cannot shuffle the order of the original array elements.

So for the movement of the elements,If the brain rotation is not very smart,It is recommended to draw a picture yourself to try it out,See how to move.

You see how these scrambled zeros move to the back

Move zero to the back,Then you need to move the non-zero elements to the front.

一种想法,我们可以用两个指针,Then start to initialize the head of the array pointing to the same time.

And how to move it？The yellow pointer starts to move,移动到3,它不是零,We replace it with the white one0的值.

继续,到这里的时候,Our yellow pointer points to zero,We certainly don't move forward zero.

There is no movement here,Do not move the white pointer,The yellow ones continue to move.
移动到4的时候,Then the number at the yellow point replaces the value at the white execution.

然后两个指针继续移动,The value at the yellow pointer now replaces the value at the white pointer again,The yellow pointer reaches the end

Then we continue to move the white pointer to the end point,and assigns the value at the following index0.

This moves it.

We write code according to this logic

class Solution {
    
    public void moveZeroes(int[] nums) {
    
        
        int pre01 =0;
        for (int pre02 = 0; pre02 < nums.length; pre02++) {
    
           if (nums[pre02]!=0)
           {
    
               nums[pre01++]=nums[pre02];
           }

        }


        for (int i = pre01; i < nums.length; i++) {
    
            nums[i]=0;

        }

    }
}

还有一种方法,There is a difference in thinking.This time we initialize two pointers.
Only this time around,我们是这样做的.

We exchange directly when comparing.

然后移动,Both are zeros here,不交换.yellow move,White unchanged

Swap next

继续

Compare carefully,There is a difference between the two methods.Our exchange patterns are different,The position to which the pointer moves is different.
来看代码实现.

  int pre01 =0;
        for (int pre02 = 0; pre02 < nums.length; pre02++) {
    
            if(nums[pre01]==0&&nums[pre02]!=0)
            {
    
               int n= nums[pre02];
               nums[pre02]=nums[pre01];
               nums[pre01++] =n;


            }
            else if (nums[pre01]!=0)
            {
    
                pre01++;
            }

        }

This execution efficiency is not as high as our above method,But the last one seems to be more like a loophole.