题目要求

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 = 199Note: 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.

思路和代码

这里涉及一个广度优先遍历，首先尝试所有合适的第一个数字和第二个数字，然后在此基础上继续判断后序的字符串是否是一个Additive Number。

为了减少无效遍历，我们可以在寻找第一个数字和第二个数字的时候及时终止。我们可以知道第一个数字的长度不应该超过字符串长度的一般，第二个数字的长度无法超过字符串长度减去第一个数字的长度。

还有一个特殊情况是以0为开头。这里我们只有当取0时是合法的。因此一旦遇到0，在判断完0作为加数时是否合法后，直接跳出循环。

public boolean isAdditiveNumber(String num) {        if(num==null || num.length() < 3) return false;        for(int i = 1 ; i<=num.length()/2; i++){            //如果以0开头，则只能取0作为加数            if(num.charAt(0)=='0' && i>1) break;            String s1 = num.substring(0, i);            long num1 = Long.parseLong(s1);            for(int j = i+1 ; j<=num.length()-i ; j++){                //如果以0开头，则只能取0作为加数                if(num.charAt(i)=='0' && j>i+1) break;                String s2 = num.substring(i, j);                long num2 = Long.parseLong(s2);                //递归判断                if(isAdditiveNumber(num.substring(j), num1, num2)){                    return true;                }            }        }        return false;    }            private boolean isAdditiveNumber(String num, long num1, long num2){        //如果剩余长度为0，说明之前都是AdditiveNumber，返回true        if(num.length()==0) return true;        long add = num1 + num2;        String adds = add + "";        //递归判断        return num.startsWith(adds) && isAdditiveNumber(num.substring(adds.length()), num2, add);    }