LeetCode26. Remove Duplicates from Sorted Array

最新推荐文章于 2024-12-06 07:02:43 发布

原创最新推荐文章于 2024-12-06 07:02:43 发布 · 159 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#c++ #leetcode

leetcode 专栏收录该内容

51 篇文章

订阅专栏

本文介绍了解决LeetCode 26题的方法，即如何在不使用额外空间的情况下从已排序数组中删除重复元素，并保持原有元素顺序不变。文章提供了C++实现的代码示例，通过遍历数组并比较相邻元素来实现去重。

LeetCode26. Remove Duplicates from Sorted Array

题目：

Given a sorted array, remove the duplicates in-place such that each element appear only once 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.

Example:

Given 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.

题目分析：判断前面的那个是否重复，如果重复，则size++。

for (int i = 1; i < nums.size(); i++) {
	if (nums[i] != nums[i - 1]) {
		nums[len++] = nums[i];
	}
}

代码：

class Solution {
public:
	int removeDuplicates(vector<int>& nums) {
		if (nums.size() <= 1) {
			return nums.size();
		}
		int len = 1;
		for (int i = 1; i < nums.size(); i++) {
			if (nums[i] != nums[i - 1]) {
				nums[len++] = nums[i];
			}
		}
		return len;
	}
};