zoukankan      html  css  js  c++  java
  • 371. Sum of Two Integers

    题目

    Calculate the sum of two integers a and b, but you are not allowed to use the operator + and -.

    Example:
    Given a = 1 and b = 2, return 3.

    分析

    不用+ - 号求两个整数的和,用二进制运算(&, |, ~, ^)

    解答

    解法1:(我)(0ms)

    public class Solution {
        public int getSum(int a, int b) {
            int c = a;
            int d = b;
            while ((c & d) != 0){
                int cc = c;
                c = c ^ d;
                d = (cc & d) << 1;
            }
            return (c ^ d);
    
        }
    }
    


    解法2+:两种方法(递归、迭代),三种运算(加法、减法、相反数)

    >"&" AND operation, for example, 2 (0010) & 7 (0111) => 2 (0010) "^" XOR operation, for example, 2 (0010) ^ 7 (0111) => 5 (0101) "~" NOT operation, for example, ~2(0010) => -3 (1101) (参与运算的都是以补码的形式)

    In bit representation, a = 0001, b = 0011,

    First, we can use "and"("&") operation between a and b to find a carry.

    carry = a & b, then carry = 0001

    Second, we can use "xor" ("^") operation between a and b to find the different bit, and assign it to a,

    Then, we shift carry one position left and assign it to b, b = 0010.

    Iterate until there is no carry (or b == 0)

    // Iterative
    public int getSum(int a, int b) {
        if (a == 0) return b;
        if (b == 0) return a;
    
        while (b != 0) {
            int carry = a & b;
            a = a ^ b;
            b = carry << 1;
        }
        
        return a;
    }
    
    // Iterative
    public int getSubtract(int a, int b) {
        while (b != 0) {
            int borrow = (~a) & b;
            a = a ^ b;
            b = borrow << 1;
        }
        
        return a;
    }
    
    // Recursive
    public int getSum(int a, int b) {
        return (b == 0) ? a : getSum(a ^ b, (a & b) << 1);
    }
    
    // Recursive
    public int getSubtract(int a, int b) {
        return (b == 0) ? a : getSubtract(a ^ b, (~a & b) << 1);
    }
    
    // Get negative number
    public int negate(int x) {
        return ~x + 1;
    }
    
  • 相关阅读:
    Python基础篇 -- 列表
    Python基础篇 -- 字符串
    Python基础篇 -- if while 语句
    Python基础篇 -- 运算符和编码
    Python 入门基础
    Docker知识收藏
    秒表
    Emac
    Android开发
    shell 小工具
  • 原文地址:https://www.cnblogs.com/xuehaoyue/p/6412662.html
Copyright © 2011-2022 走看看