26. Remove Duplicates from Sorted Array

最新推荐文章于 2024-01-02 23:35:28 发布

dici5041

最新推荐文章于 2024-01-02 23:35:28 发布

阅读量116

点赞数

原文链接：http://www.cnblogs.com/wentiliangkaihua/p/11306206.html

版权

本文介绍了一种在原地修改已排序数组以去除重复元素的方法，使用快慢指针技巧实现，确保每个元素仅出现一次，并返回新数组的有效长度。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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.

Clarification:

Confused why the returned value is an integer but your answer is an array?

Note that the input array is passed in by reference, which means modification to the input array will be known to the caller as well.

Internally you can think of this:

// nums is passed in by reference. (i.e., without making a copy)
int len = removeDuplicates(nums);

// any modification to nums in your function would be known by the caller.
// using the length returned by your function, it prints the first len elements.
for (int i = 0; i < len; i++) {
    print(nums[i]);
}
啥意思呢？简单来说就是把一个sorted array，去掉里面重复的（假装）返回不重复长度。
那怎么假装呢？让所有不重复的数先在array里面依次排列，然后长度l确定后，只打印这前l个数即可。

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