Leecode #27 Remove Element

一、 问题描述

Leecode第二十七题，题目为：

Given an array nums and a value val, 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 by modifying the input array in-place with O(1) extra memory.

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

Example 1:

Given nums = [3,2,2,3], val = 3,

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

It doesn’t matter what you leave beyond the returned length.
Example 2:

Given nums = [0,1,2,2,3,0,4,2], val = 2,

Your function should return length = 5, with the first five elements of nums containing 0, 1, 3, 0, and 4.

Note that the order of those five elements can be arbitrary.

It doesn’t matter what values are set beyond the returned length.
Clarification:

Confused why the returned value is an integer but your answer is an array?

Note that the input array is passed in by reference, which means modification to the input array will be known to the caller as well.

Internally you can think of this:

// nums is passed in by reference. (i.e., without making a copy)
int len = removeElement(nums, val);

// any modification to nums in your function would be known by the caller.
// using the length returned by your function, it prints the first len elements.
for (int i = 0; i < len; i++) {
print(nums[i]);
}

问题理解为：

给定一个数组和值val，删除与该值相等的所有元素并返回数组的新长度。
不要为另一个数组分配额外的空间，您必须使用O(1)额外内存修改输入数组。
元素的顺序可以改变。新的长度之外什么都不重要。

例1:
给定nums = [3,2,2,3]， val = 3，
函数应该返回length = 2, nums的前两个元素为2。
除去返回长度以外的返回值并不重要。

例2:
给定nums = [0,1,2,2,3,0,4,2]， val = 2，
函数应该返回length = 5, nums的前五个元素包含0、1、3、0和4。
注意，这五个元素的顺序可以是任意的。
除去返回长度以外的返回值并不重要。

说明:

为什么返回的值是整数而您的答案是数组?
注意，输入数组是通过引用传入的，这意味着调用者也知道对输入数组的修改。
你可以这样想:
// nums是通过引用传入的。(即。len = removeElement(nums, val);
//函数中对nums的任何修改都会被调用者知道。
//使用函数返回的长度，它输出第一个len元素。
for (int i = 0;i<len;i+ +){
print (num[i]);
}

二、算法思路
1、
2、

三、实现代码

class Solution {
public:
    int removeElement(vector<int>& nums, int val) {
        int res = 0;
        for (int i = 0; i < nums.size(); ++i)
        
        {
            if (nums[i] != val) 
                
                res++;
            
                nums[res] = nums[i];
        }
        
        return res;
        
    }
};