# Additive Number

### 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.

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?

### 代码​

// Additive Number// 多入口深搜// 时间复杂度O(n^3)，空间复杂度O(1)public class Solution {    public boolean isAdditiveNumber(String num) {        for (int i = 1; i <= num.length() / 2; ++i) {            if (num.charAt(0) == '0' && i > 1) continue;            for (int j = i + 1; j < num.length(); ++j) {                if (num.charAt(i) == '0' && j - i > 1) continue;                if (dfs(num, 0, i, j)) return true;            }        }        return false;    }    // 判断从 [i, j) 和 [j, k) 出发,能否走到尽头    private static boolean dfs(String num, int i, int j, int k) {        long num1 = Long.parseLong(num.substring(i, j));        long num2 = Long.parseLong(num.substring(j, k));        final String addition = String.valueOf(num1 + num2);        if (!num.substring(k).startsWith(addition)) return false;        if (k + addition.length() == num.length()) return true;        return dfs(num, j, k, k + addition.length());    }}