[LeetCode306] Additive Number

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本文介绍了一种判断字符串是否为加法数的方法。加法数是一种特殊字符串，其数字可以形成加法序列，且序列中每个数都是前两个数之和。文章详细解释了有效加法数序列的构成条件，并提供了一个具体的实现方案。

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?

Solution:

brute force

public class Solution {
    public boolean isAdditiveNumber(String num) {
        int len = num.length();
        if(len<=1) return false;
        for(int i=1;i<=len/2;i++) {
            for(int j=1;Math.max(i,j)<=len-i-j;j++) {
                if(isValid(i,j,num)) return true;
            }
        }
        return false;
    }
    public boolean isValid(int i, int j, String num) {
        if(num.charAt(i) == '0' && j>1) return false;
        if(num.charAt(0) == '0' && i>1) return false;
        Long x1 = Long.parseLong(num.substring(0,i));
        Long x2 = Long.parseLong(num.substring(i,i+j));
        String sum = "";
        for(int k=i+j; k!=num.length();k+=sum.length()) {
            x2 = x2+ x1;
            x1 = x2 - x1;
            sum = x2.toString();
            if(!num.startsWith(sum, k)) return false;
        }
        return true;
    }
}