306.Additive Number

最新推荐文章于 2025-03-22 19:42:57 发布

bo o ya ka

最新推荐文章于 2025-03-22 19:42:57 发布

阅读量86

点赞数

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/bernieloveslife/p/9772276.html

本文介绍了一种算法，用于判断一个由数字组成的字符串是否为加法数。加法数是指该字符串的数字可以形成一个至少包含三个数的加法序列，其中除了前两个数外，每个后续数都是其前两个数的和。文章通过实例解析了算法的工作原理，并讨论了处理大整数溢出的问题。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Additive number is a string whose digits can form additive sequence.

A valid additive sequence should contain at least three numbers. Except for the first two numbers, each subsequent number in the sequence must be the sum of the preceding two.

Given a string containing only digits '0'-'9', write a function to determine if it's an additive number.

Note: Numbers in the additive sequence cannot have leading zeros, so sequence 1, 2, 03 or 1, 02, 3 is invalid.

Example 1:

Input: "112358"
Output: true
Explanation: The digits can form an additive sequence: 1, 1, 2, 3, 5, 8.
1 + 1 = 2, 1 + 2 = 3, 2 + 3 = 5, 3 + 5 = 8

Example 2:

Input: "199100199"
Output: true
Explanation: The additive sequence is: 1, 99, 100, 199.
1 + 99 = 100, 99 + 100 = 199

Follow up:
How would you handle overflow for very large input integers?

class Solution:
    def isAdditiveNumber(self, num):
        """
        :type num: str
        :rtype: bool
        """
        if len(num)<=2:
            return False
        def judge(s): #judge a string valid or not
            if len(s)==1:
                return True
            if len(s)==0:
                return False
            if s[0]=='0':
                return False
            return True
        def solve(i,j): #judge this divided method vaild or not
            a,b = num[:i],num[i:j]
            if judge(a) is False or judge(b) is False:
                return False
            n1,n2 = int(a),int(b)
            # print(n1,n2)
            while True:
                n3 = n1 + n2
                c = str(n3)
                if j+len(c)>len(num) or n3!=int(num[j:j+len(c)]):
                    return False
                if j+len(c)==len(num):
                    return True
                n1 = n2
                n2 = n3
                j = j+len(c)
                # print('n2:',n2,'n1:',n1,'j:',j)
        for i in range(1,len(num)//2+1):
            for j in range(i+1,len(num)):
                # print(i,j)
                if solve(i,j):
                    return True
        return False

手动dfs 对于任一种情况，只要定下来前两个数字，后边就都是确定的了，所以只需要遍历前两个数字的所有情况。

转载于:https://www.cnblogs.com/bernieloveslife/p/9772276.html