leetcode [306]Additive Number

Posted 2021-12-01 xiaobaituyun

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：

判断字符串是否可以被划分成一个一个字符串字串，满足加法性质。

例如："112358"，可以被划分为 1, 1, 2, 3, 5, 8. 满足1 + 1 = 2, 1 + 2 = 3, 2 + 3 = 5, 3 + 5 = 8条件。

解法：

首先如果一个字符串满足这样的加法性质，那么只要确定前两个数字，那么后面的数字都会自动被确定，所以只需要确定前两个数字，就可以得到这个字符串的这个划分是否满足要求。题目要求有大数要求，所以使用BigInteger类。

import java.math.BigInteger;

class Solution {
    private boolean isVaild(BigInteger x1,BigInteger x2,int start,String num){
        if (start==num.length()) return true;
        x2=x1.add(x2);
        x1=x2.subtract(x1);
        String res=x2.toString();
        return num.startsWith(res,start)&&isVaild(x1,x2,start+res.length(),num);
    }

    public boolean isAdditiveNumber(String num) {
        int n=num.length();
        for (int i=1;i<=num.length()/2;i++){
            if (num.charAt(0)==‘0‘&&i>1) return false;
            BigInteger x1=new BigInteger(num.substring(0,i));
            for (int j=1;Math.max(i,j)<=n-i-j;j++){
                if (num.charAt(i)==‘0‘&&j>1) break;
                BigInteger x2=new BigInteger(num.substring(i,i+j));
                if (isVaild(x1,x2,i+j,num)){
                    return true;
                }
            }
        }
        return false;
    }
}