Java实现-Find the Duplicate Number

最新推荐文章于 2021-09-07 10:25:55 发布

Narasimha_Karumanchi

最新推荐文章于 2021-09-07 10:25:55 发布

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分类专栏：斩杀LintCode All in One LintCode 文章标签： LintCode Find the Duplicate N

本文链接：https://blog.youkuaiyun.com/qq_14927217/article/details/73500528

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本文介绍了一种在不修改原始数组的前提下，查找包含n+1个整数的数组中的重复数字的方法。该方法仅使用常量级额外空间，并确保时间复杂度低于O(n^2)。通过排序数组并遍历检查相邻元素来实现。

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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 number must exist. Assume that there is only one duplicate number, find the duplicate one.

注意事项

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(n^2).
There is only one duplicate number in the array, but it could be repeated more than once.

您在真实的面试中是否遇到过这个题？

Yes

样例

Given nums = [5,5,4,3,2,1] return 5
Given nums = [5,4,4,3,2,1] return 4

public class Solution {
    /**
     * @param nums an array containing n + 1 integers which is between 1 and n
     * @return the duplicate one
     */
    public int findDuplicate(int[] nums) {
        // Write your code here
        if(nums.length<2){
			return -1;
		}
		Arrays.sort(nums);
		for(int i=1;i<nums.length;i++){
			if(nums[i]==nums[i-1])
				return nums[i];
		}
		return -1;
    }
}