Given an array nums containing n + 1 integers where each integer is between 1 and n (inclusive), prove that at least one duplicate element must exist. Assume that there is only one duplicate number, find the duplicate one.
Note:
- You must not modify the array (assume the array is read only).
- You must use only constant extra space.
- Your runtime complexity should be less than
O(n2)
.
Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.
这道题给了我们n+1个数,所有的数都在[1, n]区域内,首先让我们证明必定会有一个重复数,这不禁让我想起了小学华罗庚奥数中的抽屉原理(又叫鸽巢原理), 即如果有十个苹果放到九个抽屉里,如果苹果全在抽屉里,则至少有一个抽屉里有两个苹果,这里就不证明了,直接来做题吧。题目要求我们不能改变原数组,即不能给原数组排序,又不能用多余空间,那么哈希表神马的也就不用考虑了,又说时间小于O(n2),也就不能用brute force的方法,那我们也就只能考虑用二分搜索法了,我们在区别[1, n]中搜索,首先求出中点mid,然后遍历整个数组,统计所有小于等于mid的数的个数,如果个数大于mid,则说明重复值在[mid+1, n]之间,反之,重复值应在[1, mid-1]之间,然后依次类推,知道搜索完成,此时的low就是我们要求的重复值,参见代码如下:
class Solution { public: int findDuplicate(vector<int>& nums) { int low = 1, high = nums.size() - 1; while (low <= high) { int mid = low + (high - low) * 0.5; int cnt = 0; for (auto a : nums) { if (a <= mid) ++cnt; } if (cnt <= mid) low = mid + 1; else high = mid - 1; } return low; } };
参考资料:
https://leetcode.com/discuss/60830/python-solution-explanation-without-changing-input-array
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