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.
Example 1:
Input: [1,3,4,2,2]
Output: 2
Example 2:
Input: [3,1,3,4,2]
Output: 3
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.
Solution
Language: C++
class Solution {
public:
int findDuplicate(vector<int> &nums) {
int left = 0, right = nums.size();
while (left < right) {
int mid = left + (right - left) / 2;
int count = 0;
for (auto &num : nums)
if (num <= mid)
++count;
if (count <= mid) // 加起来是合理的
left = mid + 1;
else
right = mid; // 不用减一
}
return right;
}
};
287. Find the Duplicate Number
Difficulty: Medium
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.
Example 1:
Example 2:
Note:
Solution
Language: C++