LeetCode 27.Remove Element

Description:

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 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.

分析：

太简单的一道题，可以采用vector的erase方法，遍历循环一次，将数组中与val相等的数erase掉，直接返回数组的大小即可。
注意erase方法不要出现野指针的问题。
vector标准库函数erase的声明：iterator erase (const_iterator position);，注意到形参和返回值都是迭代器。

代码如下：

#include <iostream>
#include <vector>
using namespace std;
class Solution {
public:
	int removeElement(vector<int>& nums, int val) {
		for (vector<int>::iterator iter = nums.begin(); iter != nums.end();) {
			if (*iter == val) {
				iter = nums.erase(iter);
			}
			else {
				iter++;
			}
		}
		return nums.size();
	}
};
int main() {
	Solution s;
	vector<int> nums;
	int n;
	cin >> n;
	int t;
	for (int i = 0; i < n; i++) {
		cin >> t;
		nums.push_back(t);
	}
	int val;
	cin >> val;
	cout << s.removeElement(nums, val) << endl;
	return 0;
}