01 看题和准备 今天介绍的是LeetCode算法题中Easy级别的第104题(顺位题号是461)。两个整数之间的汉明距离是相应位不同的位置数。给定两个整数x和y,计算汉明距离。 注意:0≤x,y <2^31。 例如: 输入:x = 1,y = 4 输出:2 说明: 1(0 0 0 1) 4(0 1 0 0) ↑ ↑ 上述箭头指向相应位不同的位置。
题目地址: https://leetcode.com/problems/hamming-distance/Total Accepted: 12155 Total Submissions: 16696 Difficulty: Easy题目描述The Hamming distance between two integers is the number of positions at which the corresponding bits are different.
Hamming Code Java: import java.util.Scanner;public class IntellipaatHammingDistance { public static void main(String[] args) { Scanner scanner = new Scanner(System.in); System.out.print("Enter the first string: "); String string1 = scanner.nextLine(); System.out.print("Enter the second st...
LeetCode Top 100 Liked Questions 461. Hamming Distance (Java版; Easy) 题目描述 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 < 2^31. Example: ...
Given two integersxandy, calculate the Hamming distance. PS:求海明距离。 思路:就是求x和y二进制的异或中的1的个数 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 publicclassSolution { publicinthammingDistance(intx,inty) { // String x1=Integer.toBinaryString(x); ...
原题链接在这里:https://leetcode.com/problems/total-hamming-distance/ 题目: TheHamming distancebetween 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. ...
利用Number of 1 Bits的数xor 结果的bit 1个数. 或者用Integer.bitCount() function. Time Complexity: O(1). Space: O(1). AC Java: 1publicclassSolution {2publicinthammingDistance(intx,inty) {3returnInteger.bitCount(x^y);4}5} 跟上Total Hamming Distance....
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