【重点】【二分查找】287.寻找重复数

题目
2种方法，都要理解掌握！最简单的基于HashSet的就不写在下面了，过于简单了。
一道类似的 题目，对比学习！0～n-1中缺失的数字

法1：二分查找，O(nlogn)

时间复杂度：O(nlogn)

Python

class Solution:
    def findDuplicate(self, nums: List[int]) -> int:
        l, r = 1, len(nums) - 1 # l和r分别是起始数字1和n, 不是索引
        while l < r:
            mid = (l + r) // 2
            count = 0
            for i in nums:
                if i <= mid:
                    count += 1
            if count <= mid: # 这种情况重复的数字一定大于mid!
                l = mid + 1
            else:            # 有可能重复的数字是mid
                r = mid

        return l

Java

必须熟练掌握！

class Solution {
    public int findDuplicate(int[] nums) {
        int n = nums.length;
        int left = 1, right = n; // 1和n是数字，与数组索引无关
        while (left < right) {
            int mid = left + (right - left) / 2;
            int count = 0;
            for (int i : nums) {
                if (i <= mid) {
                    ++count;
                }
            }
            if (count <= mid) { // 这种情况重复的数字一定大于mid!
                left = mid + 1;
            } else {            // 有可能重复的数字是mid
                right = mid;
            }
        }

        return left;
    }
}

法2：环形链表

参考答案

class Solution {
    public int findDuplicate(int[] nums) {
        int slow = 0, fast = 0;
        do {
            slow = next(nums, slow);
            fast = next(nums, next(nums, fast));
        } while (slow != fast);
        slow = 0;
        while (slow != fast) {
            slow = next(nums, slow);
            fast = next(nums, fast);
        }

        return slow;
    }

    public int next(int[] nums, int inx) {
        return nums[inx];
    }
}