leetcode 27. Remove Element 尾部双指针

阳光穿透心脏的1/2处 2022-06-09 08:55 143阅读 0赞

Given an array and a value, remove all instances of that value in place and return the new length.

Do not allocate extra space for another array, you must do this in place with constant memory.

The order of elements can be changed. It doesn’t matter what you leave beyond the new length.

Example:  
Given input array nums = \[3,2,2,3\], val = 3

Your function should return length = 2, with the first two elements of nums being 2.

这道题要收去除指定的元素，和上一道题类似，但是不一样，我这里直接使用一个尾部指针，不断的**把要删除的元素交换到末尾**即可完成本题。

代码如下：

/*
     * 这道题的想法就是设置一个end指针，然后
     * 
     * */
    public class Solution 
    {
        public int removeElement(int[] nums, int val) 
        {
            if(nums==null || nums.length<=0)
                return 0;
            int end=nums.length-1;
            for(int i=nums.length-1;i>=0;i--)
            {
                if(nums[i]==val)
                {
                    int tmp=nums[i];
                    nums[i]=nums[end];
                    nums[end]=tmp;
                    end--;
                }
            }
            return end+1;
        }
    
    }

下面是C++的做法

代码如下：

#include <iostream>
    #include <vector>
    using namespace std;
    
    class Solution 
    {
    public:
        int removeElement(vector<int>& nums, int val) 
        {
            if (nums.size() <= 0)
                return 0;
    
            int end = nums.size() - 1;
            for (int i = nums.size() - 1; i >= 0; i--)
            {
                if (nums[i] == val)
                {
                    int tmp = nums[i];
                    nums[i] = nums[end];
                    nums[end] = tmp;
                    end--;
                }
            }
            return end + 1;
        }
    };