Leetcode 刷题笔记之 Find the Duplicate Number

最新推荐文章于 2020-06-06 21:29:01 发布

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最新推荐文章于 2020-06-06 21:29:01 发布

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本文介绍了一种在不修改原始数组且仅使用常数级额外空间的情况下查找数组中重复数字的方法。该方法采用类似二分查找的技术，通过比较小于中间值的元素数量与预期数量来缩小搜索范围。

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题目如下：

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.

Note:

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

解答如下：

因为说要O(1) 额外空间所以还挺麻烦的。具体思路是仿照二分法，如果比mid小的数多于n/2个，那重复数一定在前半部分，否则在后半部分。代码：

class Solution {
    public int findDuplicate(int[] nums) {
        int min=0;int max=nums.length-1;
        while(min<=max){
            int mid=(min+max)/2;
            int count=0;
            for(int i=0;i<nums.length;i++){
                if(nums[i]<=mid) count++;
            }
            if(count>mid) max=mid-1;
            else min=mid+1;
        }
        return min;
    }
}