26. Remove Duplicates from Sorted Array

最新推荐文章于 2022-04-06 16:22:53 发布

原创最新推荐文章于 2022-04-06 16:22:53 发布 · 284 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#two pointers

LeetCode 专栏收录该内容

13 篇文章

订阅专栏

本文介绍了一个高效的算法，用于去除已排序数组中的重复元素，通过原地修改输入数组，确保每个元素只出现一次，并返回新的长度。算法不使用额外的空间，仅使用O(1)的额外内存，展示了两个实例，分别处理了包含少量重复项和大量重复项的数组。

Remove Duplicates from Sorted Array

【题目】

Given a sorted array nums, 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 1:

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 returned length.

Example 2:

Given nums = [0,0,1,1,1,2,2,3,3,4],

Your function should return length = 5, with the first five elements of nums being modified to 0, 1, 2, 3, and 4 respectively.

It doesn't matter what values are set beyond the returned length.

【实现】

class Solution {
    public int removeDuplicates(int[] nums) {
        if (nums == null || nums.length < 1) return 0;
        int n = nums.length;
        int j = 0;
        for (int i = 0; i < n; i++) {
            if (nums[j] != nums[i]) {
                nums[++j] = nums[i];
            }
        }
        return j + 1;
    }
}