Leecode刷题之旅-C语言/python-349两整数之和

/*
 * @lc app=leetcode.cn id=371 lang=c
 *
 * [371] 两整数之和
 *
 * https://leetcode-cn.com/problems/sum-of-two-integers/description/
 *
 * algorithms
 * Easy (55.04%)
 * Total Accepted:    8.1K
 * Total Submissions: 14.8K
 * Testcase Example:  '1\n2'
 *
 * 不使用运算符 + 和 - ​​​​​​​，计算两整数 ​​​​​​​a 、b ​​​​​​​之和。
 * 
 * 示例 1:
 * 
 * 输入: a = 1, b = 2
 * 输出: 3
 * 
 * 
 * 示例 2:
 * 
 * 输入: a = -2, b = 3
 * 输出: 1
 * 
 */
int getSum(int a, int b) {
    if(a && b) return getSum(a^b, (a&b) << 1); 
    else return a|b; 
}

这里对a和b进行二进制上的相加，然后递归中处理进位。

（不过这里一直会溢出。。。。。。尴尬）

-------------------------------------------------------------------------

python：

#
# @lc app=leetcode.cn id=371 lang=python3
#
# [371] 两整数之和
#
# https://leetcode-cn.com/problems/sum-of-two-integers/description/
#
# algorithms
# Easy (55.04%)
# Total Accepted:    8.1K
# Total Submissions: 14.8K
# Testcase Example:  '1\n2'
#
# 不使用运算符 + 和 - ​​​​​​​，计算两整数 ​​​​​​​a 、b ​​​​​​​之和。
# 
# 示例 1:
# 
# 输入: a = 1, b = 2
# 输出: 3
# 
# 
# 示例 2:
# 
# 输入: a = -2, b = 3
# 输出: 1
# 
#
class Solution:
    def getSum(self, a: int, b: int) -> int:
        while b != 0:
            carry = a & b
            a = (a ^ b) % 0x100000000
            b = (carry << 1) % 0x100000000
        return a if a <= 0x7FFFFFFF else a | (~0x100000000+1)

这里模拟32位的int 左移位，python左移位是不会溢出的。