Find the Duplicate Number

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题目：

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.

解析：

没做出来，用2分法，当小于等于n/2的数字个数大于n/2，说明多余的数字在小的部分，否则在大的部分，时间复杂度是O(nlogn)

 1 class Solution {
 2 public:
 3     int findDuplicate(vector<int>& nums) {
 4     int n=nums.size()-1;
 5     int low=1;
 6     int high=n;
 7     int mid;
 8     while(low<high){
 9         mid=(low+high)/2;
10         int count=0;
11         for(int num:nums){
12             if(num<=mid) count++;
13         }
14         if(count>mid) high=mid;
15         else low=mid+1; 
16     }
17     return low;
18     }
19 };

Find the Duplicate Number

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原文地址：http://www.cnblogs.com/raichen/p/4968229.html