The Hamming distance between two integers is the number of positions at which the corresponding bits are different. Now your job is to find the total Hamming distance between all pairs of the given numbers. Example: Input: 4, 14, 2 Output: 6 Explanat…

传送门 Description The Hamming distance between two integers is the number of positions at which the corresponding bits are different. Now your job is to find the total Hamming distance between all pairs of the given numbers. Example: Input: 4, 14, 2 Ou…

Fun one.. the punch line of this problem is quite common in Bit related problems on HackerRank - visualize it in your mind, and you will find: all bits on the same index among all numbers, will not involve other bits on other indices. So, we simply c…

The Hamming distance between two integers is the number of positions at which the corresponding bits are different. Now your job is to find the total Hamming distance between all pairs of the given numbers. Example: Input: 4, 14, 2 Output: 6 Explanat…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 位运算 日期 题目地址:https://leetcode.com/problems/total-hamming-distance/description/ 题目描述 The Hamming distance between two integers is the number of positions at which the correspondi…

［抄题］: The Hamming distance between two integers is the number of positions at which the corresponding bits are different. Now your job is to find the total Hamming distance between all pairs of the given numbers. Example: Input: 4, 14, 2 Output: 6 Ex…

题目: The Hamming distance between two integers is the number of positions at which the corresponding bits are different. Given two integers x and y, calculate the Hamming distance. Note:0 ≤ x, y < 231. Example: Input: x = 1, y = 4 Output: 2 Explanatio…

两个整数的 汉明距离 指的是这两个数字的二进制数对应位不同的数量.计算一个数组中,任意两个数之间汉明距离的总和.示例:输入: 4, 14, 2输出: 6解释: 在二进制表示中,4表示为0100,14表示为1110,2表示为0010.(这样表示是为了体现后四位之间关系)所以答案为:HammingDistance(4, 14) + HammingDistance(4, 2) + HammingDistance(14, 2) = 2 + 2 + 2 = 6.注意:    数组中元素的范围为从 0到 1…

▶ 与 Hamming  距离相关的两道题. ▶ 461. 求两个数 x 与 y 的哈夫曼距离. ● 代码,4 ms,对 x 和 y 使用异或,然后求值为 1 的位的个数. class Solution { public: int hammingDistance(int x, int y) { int i,temp; , temp = x ^ y; temp > ; temp &= temp - , i++); return i; } }; ● 代码,7 ms,转化为 bitset 来逐位比…

class Solution { public: int totalHammingDistance(vector<int>& nums) { ; ; i < ; i++) { ; for (int n : nums) { << i)) { ones++; } } res += ones * (nums.size()-ones); } return res; } };…