1. 创建中间变量 这是最快也是最简单的办法,例如: #include<stdio.h> int main(){ int a=10; int b=20; int temp; printf("交换前a,b的值为:\n"); printf("a=%d\n",a); printf("b=%d\n",b); temp=a; b=a; a=temp; printf("交换后a,b的值为:\n"); printf("…
Given a non-empty array of numbers, a0, a1, a2, … , an-1, where 0 ≤ ai < 231. Find the maximum result of ai XOR aj, where 0 ≤ i, j < n. Could you do this in O(n) runtime? Example: Input: [3, 10, 5, 25, 2, 8] Output: 28 Explanation: The maximum resul…
You are given two linked lists representing two non-negative numbers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list. Input: (2 -> 4 -> 3) + (5 -> 6 ->…
2.5 You have two numbers represented by a linked list, where each node contains a single digit. The digits are stored in reverse order, such that the 1's digit is at the head of the list. Write a function that adds the two numbers and returns the sum…
javascript在编程时经常会涉及到如何交换两个变量的值,例如常见的冒泡排序,快速排序等:下面我讲根据自己近期所学总结几种常见的交换两个变量值的方法: 方法一:借助第三方变量交换两个变量的值 var num1=20; var num2=50; var temp=num1; num1=num2; num2=temp; console.log(num1);//在控制台输出交换后的num1=50 console.log(num2);//在控制台输出交换后的num2=20 方法二:借助加法计算,交换…
C语言中要实现交换两个数的值,可以有很多种方法,具体如下所述. 不使用中间变量: // 异或, a^=b^=a^=b; a ^= b; b ^= a; a ^= b; // 加减 a = a + b; b = a - b; a = a - b; // 乘除 a = a * b; b = a / b; b = a/ b; 使用中间变量: // 需临时空间 temp = a; a = b; b = temp; 正如你所想的那样,上面所示代码只是描述了交换两个数的值的思想,在你实际使用时,还有诸多地方…
反射是很强大的,谁说的final修饰的就不能改变, 通过反射获取成员变量,之后可以取消访问修饰符,也就是说private的也可以访问, 在修改常量(final修饰的),之后就可以对其做任何操作了 如下,通过一个方法交换两个Integer对象的值: package ni.jun.yang.test; import java.lang.reflect.Field; import java.lang.reflect.Modifier; public class Test { public static…
