Given an array of integers nums containing n + 1 integers where each integer is in the range [1, n] inclusive.

There is only one repeated number in nums, return this repeated number.

You must solve the problem without modifying the array nums and uses only constant extra space.

Example 1:

Input: nums = [1,3,4,2,2]
Output: 2

Example 2:

Input: nums = [3,1,3,4,2]
Output: 3

Example 3:

Input: nums = [1,1]
Output: 1

Example 4:

Input: nums = [1,1,2]
Output: 1

Constraints:

• 1 <= n <= 105
• nums.length == n + 1
• 1 <= nums[i] <= n
• All the integers in nums appear only once except for precisely one integer which appears two or more times.

nums[nums[i]-1] *= -1， 如果已经<0 了，则证明 nums[i]是重复的数字

代码实现(Rust):

impl Solution {
    pub fn find_duplicate(mut nums: Vec<i32>) -> i32 {
        for i in 0..nums.len() {
            let idx = nums[i].abs() as usize;
            if nums[idx-1] < 0 {
                return idx as i32;
            }
            nums[idx-1] *=  -1;
        }
        unreachable!()
    }
}