dfs/additive-number

优质

小牛编辑

129浏览

2023-12-01

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?

分析

这是一个多阶段决策问题，且必须走到字符串最后一个字符才能得出结论，因此适合用深搜或DP。

再仔细想一下状态转换图，每次索引变化一下，就跟之前的完全没有重复，因此状态转换图是一颗树，不是DAG，因此不存在重叠子问题，因此排除DP，本题应该用深搜。

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());
    }
}