【LeetCode】306. Additive Number 解题报告（Python）

标签（空格分隔）： LeetCode

作者： 负雪明烛 id： fuxuemingzhu 个人博客： http://fuxuemingzhu.me/

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题目地址：https://leetcode.com/problems/additive-number/description/

题目描述：

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?

题目大意

给出了一个有0-9数字组成的纯数字字符串。判断它能不能组成所谓的“加法数字”，即费布拉奇数列。注意这个题注重点在不管你几位数字去划分，只要满足后面的数字等于前两个的和即可。

解题方法

是不是很眼熟呢？和我们刚刚做过的判断是否是有效的IP93. Restore IP Addresses如出一辙：使用回溯法+合理剪枝。

因为只要判断能否构成即可，所以不需要res数组保存结果。回溯法仍然是对剩余的数字进行切片，看该部分切片能否满足条件。剪枝的方法是判断数组是否长度超过3，如果超过那么判断是否满足费布拉奇数列的规则。不超过3或者已经满足的条件下继续进行回溯切片。最后当所有的字符串被切片完毕，要判断下数组长度是否大于等于3，这是题目要求。

代码如下：

class Solution(object):
    def isAdditiveNumber(self, num):
        """
        :type num: str
        :rtype: bool
        """
        return self.dfs(num, [])
        
    def dfs(self, num_str, path):
        if len(path) >= 3 and  path[-1] != path[-2] + path[-3]:
            return False
        if not num_str and len(path) >= 3:
            return True
        for i in range(len(num_str)):
            curr = num_str[:i+1]
            if (curr[0] == '0' and len(curr) != 1):
                continue
            if self.dfs(num_str[i+1:], path + [int(curr)]):
                return True
        return False

日期

2018 年 6 月 12 日 ———— 实验室上午放假2333刷题吧