LeetCode: Find the Duplicate Number

Posted

tags:

篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了LeetCode: Find the Duplicate Number相关的知识,希望对你有一定的参考价值。

problem:

Given an array nums containing n + 1 integers where each integer is between 1 and n (inclusive), prove that at least one duplicate number must exist. Assume that there is only one duplicate number, find the duplicate one.

Note:

  1. You must not modify the array (assume the array is read only).
  2. You must use only constant, O(1) extra space.
  3. Your runtime complexity should be less than O(n2).
  4. There is only one duplicate number in the array, but it could be repeated more than once.

Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.

Subscribe to see which companies asked this question

 

 1 class Solution {
 2 public:
 3     int findDuplicate(vector<int>& nums) {
 4         //二分查找法的运用
 5         
 6         int low=0,high=nums.size()-1;
 7         
 8            
 9         while(low<=high)
10         {
11             int cnt=0;
12             int middle=low+(high-low)/2;
13             
14             for(int i=0;i<nums.size();i++)
15             {
16                 if(nums[i]<=middle)
17                   cnt++;
18             }
19             
20             if(cnt>middle)
21                 high=middle-1;
22             else
23                 low=middle+1;
24         }
25         
26         return low;
27         
28         
29     }
30 };

 

以上是关于LeetCode: Find the Duplicate Number的主要内容,如果未能解决你的问题,请参考以下文章

LeetCode: Find the Duplicate Number

LeetCode 997. Find the Town Judge

Leetcode: Find the Difference

LeetCode 389. Find the Difference

Leetcode 389 Find the difference

LeetCode 389. Find the Difference