Code & Fun

第55天。

今天的题目是Find the Duplicate Number:

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.

一开始,没看到说要用O(1)的空间复杂度,就直接用计数的方法去写了:

1
2
3
4
5
6
7
8
int findDuplicate1(vector<int>& nums) {
vector<int> count(nums.size(),0);
for(auto i:nums) {
count[i]++;
if (count[i]>1) return i;
}
return -1;
}

然后是后来想了很久,想着利用异或的方法去做,就是先将nums中的数字进行异或,然后在对[1,n]的数字进行异或,然后就直接是答案了,但是这种问题要限定在重复数字只重复一次的情况下,即需要保证[1:n]的数字都存在:

1
2
3
4
5
6
7
8
9
10
int findDuplicate2(vector<int>& nums) {
int t = 0;
int n = nums.size() - 1;
for(auto i:nums) {
t ^= i;
}
for(int i = 1;i <= n;i++)
t ^= i;
return t;
}

最后。。。最后就实在想不出了,只好去看dicuss了:

他是用了一个List Cycle中找环点的方式,这里的链表中的nxet就是用nums的值来表示的。

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
int findDuplicate(vector<int>& nums) {
int n = nums.size();
int slow = n;
int fast = n;
do {
slow = nums[slow-1];
fast = nums[nums[fast-1]-1];
}while(slow != fast);
slow = n;
while(slow != fast) {
slow = nums[slow - 1];
fast = nums[fast - 1];
}
return slow;
}

本文首发于Code & Fun