一、问题描述: Given an array of size n, find the majority element. The majority element is the element that appears more than ⌊ n/2 ⌋ times. You may assume that the array is non-empty and the majority
题目内容 https://leetcode-cn.com/problems/majority-element/ Given an array of size n, find the majority element. The majority element is the element that appears more than ⌊ n/2 ⌋times. You may assume
要求求出给定数组中出现频率大于 n / 2的数,那么这个数显然只可能有1个。 题目为: Given an array of size n, find the majority element. The majority element is the element that appears more than ⌊ n/2 ⌋ times. You may assume that the ar
题目: Problem - 276D - Codeforces 思路: l <= r 1. l == r ,异或的结果为0 2. l < r (注意:二进制位中,某一位是1,这一位后面的全是1,也没这个1大,比如1000 > 0111) r比l大,我们找到第一个不同的位置,这个位置一定是r是1,l是0 (因为r>l). (前面的位置都不用考虑了,无法操作到。) 此时 r
Problem Given an array of size n, find the majority element. The majority element is the element that appears more than ⌊ n/2 ⌋ times. You may assume that the array is non-empty and the majority ele
题目描述:一个非空数组,里面有元素出现了数组长度一半以上,返回这个元素 解题思路: 最容易想的是排序,返回中间那个元素即可,时间复杂度O(nlog(n)),空间复杂度O(l);另一种思路借助于哈希表,遍历数组每一个元素,出现次数记录在hash表中,若hash值超过了数组长度的一半,返回这个元素即可,时间复杂度O(n),空间复杂度O(n)。 C++实现如下: 解法一:排序 class Sol