LeetCode：287. Find the Duplicate Number（找出重复的数）

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:

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 than O(n2).
4. There is only one duplicate number in the array, but it could be repeated more than once.

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方法1：（这个题目比较简单，思路也比较直接）

private int findDuplicateNumber1(int[] nums) {
        if (nums.length <= 1) {
            return -1;
        }
        Arrays.sort(nums);
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] == nums[i + 1]) {
                return nums[i];
            }
        }
        return -1;
    }

时间复杂度：O(n.logn)

空间复杂度：O(n)

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方法2：（利用set集合的形式）

private int findDuplicateNumber2(int[] nums) {
        Set<Integer> set = new HashSet<>();
        for (int num : nums) {
            if (!set.contains(num)) {
                set.add(num);
            } else {
                return num;
            }
        }
        return -1;
    }

时间复杂度：O（n）

空间复杂度：O(1)