[LeetCode]27. Remove Element

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#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值的方法，并保持了常数级别的内存消耗。通过两个不同的实现方案展示了如何高效地完成这一任务。

题目：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.

两种方法如下：

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

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