Leetcode27：Remove Element

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本文介绍了一种高效的算法，用于在不使用额外空间的情况下，从数组中移除所有指定值的实例，并返回修改后的长度。通过双指针技术，该算法实现了O(n)的时间复杂度和O(1)的空间复杂度，展示了如何在保持输入数组结构的同时完成这一任务。示例包括了具体的代码实现和效果验证。

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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.

C++
int removeElement(vector<int>& nums, int val) {
        if(nums.empty())
            return 0;
        int cur = 0;
        int pre = 0;
        int n = nums.size();
        while(cur < n)
        {
            if(nums[cur] == val)
                ++cur;
            else
                nums[pre++] = nums[cur++];
        }
        return pre;
    }