306.Additive Number

Posted 2021-01-17 bernieloveslife

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。