Google算法题：M-Find the Duplicate Number

http://www.lintcode.com/zh-cn/problem/find-the-duplicate-number/

http://www.jiuzhang.com/article/Google%202016%E5%B9%B412%E6%9C%88New%20Grad%20SDE%E7%94%B5%E9%9D%A2%E9%9D%A2%E7%BB%8F/

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.


 注意事项

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

您在真实的面试中是否遇到过这个题？ 

Yes

样例

Given 

nums

 = 

[5,5,4,3,2,1]

 return 

5

Given 

nums

 = 

[5,4,4,3,2,1]

 return 

4

public class Solution {
/**
* @param nums an array containing n + 1 integers which is between 1 and n
* @return the duplicate one
*/
//TC=O(nlgn)
public int findDuplicate(int[] nums) {
// Write your code here
if(nums==null || nums.length==0) return -1;

int n = nums.length;
int s=1, e=n-1;
while(s<=e){
int m = (s+e)/2;
int lowerCount=0, higerCount=0, euqalCount=0;
for(int i=0; i<n; i++){
if(m==nums[i]){
euqalCount++;
if(euqalCount>1)
return m;
}else if(m<nums[i]){
higerCount++;//比m大的数的计数
}else{
lowerCount++;//比m小的数的计数
}
}
if(higerCount>(n-1-m))//由于是n+1个数，多出的一个可定是重复的数
s=m+1;
else if(lowerCount>(m-1))
e=m-1;
}

return -1;
}
}