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 thanO(n2). There is only one duplicate number in the array, but it could be repeated more than once 非常好的题目,开始是用二分做的,比...
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Given an arraynumscontainingn+ 1 integers where each integer is between 1 andn(inclusive), prove that at least one duplicate number must exist. Assume that there is only one duplicate number, find the duplicate one. 给一个数组,其中一定有且只有一个数字是重复的,找出那个数字; O(n2)的算法就...
这道题给了我们一个数组,数组中的数字可能出现一次或两次,让我们找出所有出现两次的数字,由于之前做过一道类似的题目Find the Duplicate Number,所以不是完全无从下手。这类问题的一个重要条件就是1 ≤ a[i] ≤ n (n = size of array),不然很难在O(1)空间和O(n)时间内完成。首先来看一种正负替换的方法...
- There is only one duplicate number in the array, but it could be repeated more than once. 代码如下: public int findDuplicate(int[] nums) { if (nums.length > 1){ int slow = nums[0]; int fast = nums[nums[0]]; //以下为求两个指针第一次相遇的点 while (slow != fast){ slow...
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. ...
There is only one duplicate number in the array, but it could be repeated more than once 本题给定的条件是数组,数组里面有n+1个整数,其中包括1～n个数,其中有一个数是重复的,求那个重复的数。 如图所示: 此题分为两个steps 1.第一次相遇的时候,2m = m+c*r--->m=c*r@; n+a=m...
publicintfindDuplicate(int[]nums){HashSet<Integer>set=newHashSet<>();for(inti=0;i<nums.length;i++){if(set.contains(nums[i])){returnnums[i];}set.add(nums[i]);}return-1;} 解法三 二分查找 参考这里。 我们知道二分查找要求有序,但是给定的数组不是有序的,那么怎么用二分查找呢?
There is only one duplicate number in the array, but it could be repeated more than once. 解题思路 将数组的索引作为地址,存入数组的元素作为值,每个数组单元作为一个节点 由于不存在值为 0 的节点,因此构成的节点一定不会在首节点形成环 由于值中存在重复元素,因此构成的链表一定存在重复的节点指向同一个...
https://leetcode.com/problems/find-the-duplicate-number/题目找出一组无序数字中重复的数字。其中, 数组长度为 n+1,数字取值范围是 [1, n]有且只有一个数字重复要求数组不能改变(只读)只能使用 constant sp…