[LeetCode] Find the Duplicate Number 寻找重复数

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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:

1. You must not modify the array (assume the array is read only).
2. You must use only constant extra space.
3. 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

LeetCode All in One 题目讲解汇总(持续更新中...)

[LeetCode] Find the Duplicate Number 寻找重复数

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原文地址：http://www.cnblogs.com/grandyang/p/4843654.html