415. Add Strings

转载于 2016-10-15 15:30:00 发布 · 44 阅读

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CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/salmd/p/5964545.html

本文介绍了一种不使用内置大整数库或直接类型转换的方法来实现两个非负数字符串的相加操作。通过字符与数字之间的映射转换，并采用逆序遍历的方式进行逐位相加，最终返回相加后的字符串结果。

problem

Given two non-negative numbers num1 and num2 represented as string, return the sum of num1 and num2.

Note:

The length of both num1 and num2 is < 5100.
Both num1 and num2 contains only digits 0-9.
Both num1 and num2 does not contain any leading zero.
You must not use any built-in BigInteger library or convert the inputs to integer directly.

将0-9(非负数)组成的字符串，相加返回相加后的str

不能用内置的类型转换函数

solution

实现的比较笨拙，两个字典实现字符数字互相转换
这个思路其实是实现的类型转换，不是字符串相加
继续优化的话，可以将loop缩减为1个

class Solution(object):
    def addStrings(self, num1, num2):
        """
        :type num1: str
        :type num2: str
        :rtype: str
        """
        dictStoN = {'0':0,'1':1,'2':2,'3':3,'4':4,'5':5,'6':6,'7':7,'8':8,'9':9}
        dictNtoS = dict([(v,k) for k,v in dictStoN.iteritems()])
        nums = 0 
        res = ''
        #num1 = sum(map(dictNum[num1[x]]*10**x for x in range(len(num1))[::-1])
        
        n = 0
        for x in num1[::-1]:
            nums += dictStoN[x]*10**n
            n += 1
            
        n = 0
        for x in num2[::-1]:
            nums += dictStoN[x]*10**n
            n += 1
            
        if nums is not 0:    
            while 1:
                if nums < 10:
                    res += dictNtoS[nums]
                    break
                nums,mid = divmod(nums,10) 
                res += dictNtoS[mid]
        else:
            res = '0'
        
        return res[::-1]

discuss

def addStrings(self, num1, num2):
    z = itertools.izip_longest(num1[::-1], num2[::-1], fillvalue='0')
    res, carry, zero2 = [], 0, 2*ord('0')
    for i in z:
        cur_sum = ord(i[0]) + ord(i[1]) - zero2 + carry
        res.append(str(cur_sum % 10))
        carry = cur_sum // 10
    return ('1' if carry else '') + ''.join(res[::-1])

转载于:https://www.cnblogs.com/salmd/p/5964545.html