LeetCode 27. Remove Element

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本文介绍了一种在不使用额外空间的情况下从数组中移除指定元素的方法，并保持其他元素的相对顺序不变。通过两个指针的技术实现原地删除，最终返回不含目标元素的新数组长度。

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.

// Two pointers.

#include <vector>
#include <iostream>
using namespace std;

int removeElement(vector<int>& nums, int target) {
    if(nums.size() == 0) return 0;
    int i = 0;  // one pointer.
    int j = nums.size() - 1; // the other pointer.
    while(i <= j) {
        if(nums[i] == target) {
            swap(nums[i], nums[j]);
            j--;
        } else {
            i++;
        }
    }
    return i;
}

int main(void) {
    vector<int> nums;
    nums.push_back(3);
    nums.push_back(3);
    int length = removeElement(nums, 3);
    cout << length << endl;
}