Given an arraynumscontainingn+ 1 integers where each integer is between 1 andn(inclusive), prove that at least one duplicate number must exist. Assume that there is only one duplicate number, find the duplicate one.

Note:

1. You must not modify the array (assume the array is read only).
2. You must use only constant, O(1) extra space.
3. Your runtime complexity should be less thanO(n2).
4. There is only one duplicate number in the array, but it could be repeated more than once.

Solution 1: The main idea is the same with problem Linked List Cycle II. Use two pointers the fast and the slow. The fast one goes forward two steps each time, while the slow one goes only step each time. They must meet the same item when slow==fast. In fact, they meet in a circle, the duplicate number must be the entry point of the circle when visiting the array from nums[0]. Next we just need to find the entry point. We use a point(we can use the fast one before) to visit form begining with one step each time, do the same job to slow. When fast==slow, they meet at the entry point of the circle.

int findDuplicate3(vector<int>& nums)
{
    if (nums.size() > 1)
    {
        int slow = nums[0];
        int fast = nums[nums[0]];
        while (slow != fast)
        {
            slow = nums[slow];
            fast = nums[nums[fast]];
        }

        fast = 0;
        while (fast != slow)
        {
            fast = nums[fast];
            slow = nums[slow];
        }
        return slow;
    }
    return -1;
}

Solution 2: binary search: At first the search space is numbers between 1 to n. Each time I select a number mid (which is the one in the middle) and count all the numbers equal to or less than mid. Then if the count is more than mid, the search space will be [1 mid] otherwise [mid+1 n]. I do this until search space is only one number.

    public int findDuplicate(int[] nums) {
        if (nums == null || nums.length < 2) {
            return 0;
        }
        int i = 1, j = nums.length - 1;
        while (i < j) {
            int mid = (i + j) / 2, count = 0;
            for (int num : nums) {
                if (num <= mid) {
                    count++;
                }
            }
            if (count <= mid) {
                i = mid + 1;
            } else{
                j = mid;
            }
        }
        return i;
    }

Solution 3: bit manipulate: We can count the sum of each 32 bits separately for the given array (stored in "b" variable) and for the array [1, 2, 3, ..., n] (stored in "a" variable). If "b" is greater than "a", it means that duplicated number has 1 at the current bit position (otherwise, "b" couldn't be greater than "a").

public int findDuplicate(int[] nums) {
    int n = nums.length-1, res = 0;
    for (int p = 0; p < 32; ++ p) {
        int bit = (1 << p), a = 0, b = 0;
        for (int i = 0; i <= n; ++ i) {
            if (i > 0 && (i & bit) > 0) ++a;
            if ((nums[i] & bit) > 0) ++b;
        }
        if (b > a) res += bit;
    }
    return res;
}