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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:
O(n2).Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.
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思路：
（1）直接看个例子
index: 0 1 2 3 4 5 6 7 8
nums: 2 1 4 7 5 6 4 3 8
（2）从index = 0出发，可以看到有这样一个循环：
index: 0 -> 2 -> 6 -> 4 -> 6 ->...
nums[index]: 2 -> 4 -> 5 -> 6 -> 4->...
（3）可以看到从[4->5->6->4]是一个循环，正好4就是我们要求的duplicate number。
是不是有点像：Linked List Cycle II 中找环的入口点。
c++ code:
class Solution {
public:
int findDuplicate(vector<int>& nums) {
int size = nums.size();
if(size < 2) return -1;
int slow = nums[0];
int fast = nums[nums[0]];
while(slow != fast) {
slow = nums[slow];
fast = nums[nums[fast]];
}
fast = 0;
while(slow != fast) {
slow = nums[slow];
fast = nums[fast];
}
return slow;
}
};
LeetCode:Find the Duplicate Number
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原文地址：http://blog.csdn.net/itismelzp/article/details/51329627
