Remove Duplicates in Sorted Array (I /II)

本文介绍如何使用双指针技术去除有序数组中的重复元素,并确保每个元素只出现一次或两次。提供两种不同的C++实现方法,包括代码示例。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

26Remove Duplicates from Sorted Array

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.

双指针,i指向没有duplicates的字段的末尾(不包含最后一个数),j向后遍历寻找不重复的数,找到就加到i的位置上。两种代码实现:

第一种是我自己的写的:

int removeDuplicates(vector<int>& nums) {
    if (nums.size() == 0) {
        return 0;
    }
    int i = 0, j = 0, rightBound = (int)nums.size() - 1;
    nums.push_back(nums[0] - 1);
    while (j <= rightBound) {
        while (j <= rightBound && nums[j] != nums[j+1]) {
            nums[i++] = nums[j++];
        }
        while (j <= rightBound && nums[j] == nums[j+1]) {
            j++;
        }
    }
    return i;
}

第二种是discuss里面别人的做法,简洁很多:

int removeDuplicates(vector<int>& nums) {
    int i = 0;
    for (int n : nums)
        if (!i || n > nums[i-1])
            nums[i++] = n;
    return i;
}

两种里面i都指向可以直接插入的位置。


80Remove Duplicates from Sorted Array II

Given a sorted array nums, remove the duplicates in-place such that duplicates appeared at most twice 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 = [1,1,1,2,2,3],
Your function should return length = 5, with the first five elements of nums being 1, 1, 2, 2 and 3 respectively.

It doesn't matter what you leave beyond the returned length.

通用是双指针的思路,同样有两种实现,

第一种:

// Idea: two pointers
// j runs faster then i. i tracks the right-side boudary of result array, j runs ahead to move valid nums to nums[i]
// [0, 0, 1, 3, 3, 7, 7, 7, 7, 8, 9, 10, ...]
//                       i     j

int removeDuplicates(vector<int>& nums) {
    if (nums.size() == 0) {
        return 0;
    }
    int i = 0, j = 0, rightBound = (int)nums.size() - 1;
    nums.push_back(nums[0] - 1);
    while (j <= rightBound) {
        while (j <= rightBound && nums[j] != nums[j+1]) {
            nums[i++] = nums[j++];
        }
        if (j < rightBound) {
            nums[i++] = nums[j++];
        }
        while (j <= rightBound && nums[j] == nums[j+1]) {
            j++;
        }
    }
    return i;
}

第二种:

int removeDuplicates(vector<int>& nums) {
    int i = 0;
    for (int n : nums)
        if (i < 2 || n > nums[i-2])
            nums[i++] = n;
    return i;
}

评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值