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:
- 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(n2)
. - There is only one duplicate number in the array, but it could be repeated more than once.
class Solution {
//可以理解成链表
//由于值为1~n,则0位置一定指向下一个node
//一旦链表构成环,则说明环开始的节点就是重复的数字(因为目前链表中的数字都是访问过的,再次访问说明出现重复值)
//可以用快慢指针找到环开始的节点
public int findDuplicate(int[] nums) {
int slow = nums[0];//走一步
int fast = nums[nums[0]];//走两步
while(slow!=fast){
slow = nums[slow];
fast = nums[nums[fast]];
}
fast = 0;
//设起点到交点长 m, 交点到快慢指针相遇长 s,环长r
//则有到快慢指针相遇,慢指针走=m+s,快指针走2(m+s) = m+s+kr -> m = kr-s
//快慢指针剩余 r-s 步能做到交点
//当 r = 1, m = r-s
while(fast!=slow){
slow = nums[slow];
fast = nums[fast];
}
return slow;
}
}