287. Find the Duplicate Number

本文介绍了在数组中寻找重复数字的两种高效算法:鸽笼原理二分法和映射找环法。鸽笼原理二分法利用了数组中数字范围的特性,通过二分搜索确定重复数字;映射找环法则将数组转换为链表结构,利用快慢指针找到环的入口即重复数字。

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方法一:应用鸽笼原理进行二分法查找

因为在1~n的范围内最多只能放n个不同的数字,但现在数组放了n+1个数字,所以至少有一个重复。

对于在1~n范围内的某个数字m,那么数组中小于等于m的数字最少有m个,并且刚好为m的时候,1~m之间不会有重复,比m大时,重复数字必在1~m中。

public class Solution {
    public int findDuplicate(int[] nums) {
        int low = 1, high = nums.length-1;
        while (low<high) {
            int mid = (low+high)/2;
            int count = 0;
            for(int num: nums) if (num<=mid) count++;
            if (count > mid) high = mid; else low = mid + 1;
        }
        return low;
    }
}

 方法二:映射找环法

假设数组中没有重复,那我们可以做到这么一点,就是将数组的下标和1到n每一个数一对一的映射起来。比如数组是213,则映射关系为0->2, 1->1, 2->3。假设这个一对一映射关系是一个函数f(n),其中n是下标,f(n)是映射到的数。如果我们从下标为0出发,根据这个函数计算出一个值,以这个值为新的下标,再用这个函数计算,以此类推,直到下标超界。实际上可以产生一个类似链表一样的序列。比如在这个例子中有两个下标的序列,0->2->3

但如果有重复的话,这中间就会产生多对一的映射,比如数组2131,则映射关系为0->2, {1,3}->1, 2->3。这样,我们推演的序列就一定会有环路了,这里下标的序列是0->2->3->1->1->1->1->...,而环的起点就是重复的数(Floyd's Algorithm)。

考虑如下的一个序列:

从idx=0开始,对应的val是下一个idx,建立一个类似链表的结构, 1->3->2->4->2 将会出现一个环,之后就是链表判环的解法,可以用快慢指针来做。环的入口就是重复的数字。 

                                                     

class Solution {
    public int findDuplicate(int[] nums) {
        int slow = nums[0], fast = nums[0];
        
        for(;;) {
            
            slow = nums[slow];
            fast = nums[nums[fast]];
            if(slow==fast)
                break;
        }

        fast = nums[0];
        while (slow != fast) {
            slow = nums[slow];
            fast = nums[fast];
        }
        return slow;
    }
}

 

(c++题解,代码运行时间小于200ms,内存小于64MB,代码不能有注释)It’s time for the company’s annual gala! To reward employees for their hard work over the past year, PAT Company has decided to hold a lucky draw. Each employee receives one or more tickets, each of which has a unique integer printed on it. During the lucky draw, the host will perform one of the following actions: Announce a lucky number x, and the winner is then the smallest number that is greater than or equal to x. Ask a specific employee for all his/her tickets that have already won. Declare that the ticket with a specific number x wins. A ticket can win multiple times. Your job is to help the host determine the outcome of each action. Input Specification: The first line contains a positive integer N (1≤N≤10 5 ), representing the number of tickets. The next N lines each contains two parts separated by a space: an employee ID in the format PAT followed by a six-digit number (e.g., PAT202412) and an integer x (−10 9 ≤x≤10 9 ), representing the number on the ticket. Then the following line contains a positive integer Q (1≤Q≤10 5 ), representing the number of actions. The next Q lines each contain one of the following three actions: 1 x: Declare the ticket with the smallest number that is greater than or equal to x as the winner. 2 y: Ask the employee with ID y all his/her tickets that have already won. 3 x: Declare the ticket with number x as the winner. It is guaranteed that there are no more than 100 actions of the 2nd type (2 y). Output Specification: For actions of type 1 and 3, output the employee ID holding the winning ticket. If no valid ticket exists, output ERROR. For actions of type 2, if the employee ID y does not exist, output ERROR. Otherwise, output all winning ticket numbers held by this employee in the same order of input. If no ticket wins, output an empty line instead. Sample Input: 10 PAT000001 1 PAT000003 5 PAT000002 4 PAT000010 20 PAT000001 2 PAT000008 7 PAT000010 18 PAT000003 -5 PAT102030 -2000 PAT000008 15 11 1 10 2 PAT000008 2 PAT000001 3 -10 1 9999 1 -10 3 2 1 0 3 1 2 PAT000001 3 -2000 Sample Output: PAT000008 15 ERROR ERROR PAT000003 PAT000001 PAT000001 PAT000001 1 2 PAT102030
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