leetcodeOJ 306: Additive Number

题目来源：https://leetcode.com/problems/additive-number/description/

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.

For example:
“112358” is an additive number because 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
“199100199” is also an additive number, the additive sequence is: 1, 99, 100, 199.
1 + 99 = 100, 99 + 100 = 199
Note: Numbers in the additive sequence cannot have leading zeros, so sequence 1, 2, 03 or 1, 02, 3 is invalid.

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

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

为了方便处理大整数，利用java中的BigInteger类，及相关函数

代码如下：

import java.math.BigInteger;//别忘了这一句，不然会出错class Solution {    public boolean isAdditiveNumber(String num) {        int n = num.length();        for(int i = 1; i <= n/2; ++i){            if(i > 1 && num.charAt(0) == '0')                return false;            BigInteger num1 = new BigInteger(num.substring(0, i));            for(int j = 1; Math.max(i, j) <= n-i-j; ++j){//注意结束条件                if(j > 1 && num.charAt(i) == '0') //数字以0开头，类似03、02这种，要考虑在内                    break;                BigInteger num2 = new BigInteger(num.substring(i, i+j));                if(isValid(num, num1, num2, i+j))                    return true;                            }        }        return false;    }    public boolean isValid(String num, BigInteger num1, BigInteger num2, int pos){        if(pos == num.length())            return true;        num2 = num2.add(num1);        num1 = num2.subtract(num1);        String str = num2.toString();        return num.startsWith(str, pos) && isValid(num, num1, num2, pos+str.length());                              }}