leetcode 第十三周算法报告

最新推荐文章于 2025-09-01 21:45:36 发布

原创最新推荐文章于 2025-09-01 21:45:36 发布 · 223 阅读

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本文介绍了一种从已排序数组中去除重复元素的方法，确保每个元素只出现一次，并返回处理后的数组长度。示例中，输入数组 [1,1,2] 经过处理后，前两个元素为 1 和 2，新长度为 2。

Remove Duplicates from Sorted Array

Given a sorted array, remove the duplicates in place such that each element appear only once and return the new length.

For example,
Given input array nums = [1,1,2],

Your function should return length = 2, with the first two elements of nums being 1 and 2 respectively. It doesn't matter what you leave beyond the new length.

解题思路：

很简单，重新生成数组即可。

代码如下：

#include<algorithm>
class Solution{
public:
	int removeDuplicates(vector<int>& nums) {
		vector<int> res;
		if (nums.size() == 0) return 0;
		int l = 1;
		int former = nums[0];
		res.push_back(nums[0]);
		int preIdx = 0
		for (int i = 1; i < nums.size(); ++i){
			if (former != nums[i]){
				++l;
				former = nums[i];
			}
		}
		nums = res;
		return l;
	}
};

结果如下：