[Leetcode] 306. Additive Number 解题报告-优快云博客

本文链接：https://blog.youkuaiyun.com/magicbean2/article/details/75703095

本文介绍了一种算法，用于判断一个仅包含数字的字符串是否为加性数。加性数是指该字符串的数字能形成一个至少包含三个数的加性序列。文章详细解释了加性序列的概念，并提供了一个具体的实现方案，包括如何枚举并验证序列。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题目：

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?

思路：

通过分析题意可知，一旦确定了第一个数和第二个数，则后面的任务就是验证了。所以我们只需要枚举第一个和第二个数，然后采用DFS不断验证后面的字符串子串是否是前两个的和即可，思路比较简单。程序里面可以优化的部分包括：1）在DFS函数中，一旦发现num3的长度小于num1的长度或num2的长度，就可以提前返回false；2）在DFS循环验证sum的时候，str1的长度不必要从1开始取，只需要从sum的长度开始取就可以了。需要注意的特殊情况是：如果‘0’出现在某个字符串的首位，那么它就只能以"0"的形式出现，所以在循环第一遍完成之后，一旦发现某个字符串的首字符是‘0’，则立即跳出。为了使得我们的算法可以处理任意大的整数，可以用字符串相加来模拟整数相加。

代码：

class Solution {
public:
    bool isAdditiveNumber(string num) {
        if (num.length() < 3) {
            return false;
        }
        int len = num.length();
        for (int i = 1; i <= len - 2; ++i) {            // enumerate all the cases
            string num1 = num.substr(0, i);
            for (int j = i + 1; j <= len - 1; ++j) {
                string num2 = num.substr(i, j - i), num3 = num.substr(j);
                if (DFS(num1, num2, num3)) {
                    return true;
                }
                if (num[i] == '0') {
                    break;
                }
            }
            if(num[0] == '0') {
                break;
            }
        }
        return false;
    }
private:
    bool DFS(string &num1, string &num2, string &num3) {
        if (num3.length() == 0) {
            return true;
        }
        if (num3.length() < num1.length() || num3.length() < num2.length()) {
            return false;                                           // this judgement will accelerate the program a lot
        }
        string sum = getSum(num1, num2);
        for (int i = sum.length(); i <= num3.length(); ++i) {       // i do not have to start from 1
            string str1 = num3.substr(0, i), str2 = num3.substr(i);
            if (str1 == sum && DFS(num2, str1, str2)) {
                return true;
            }
            if (num3[0] == '0') {
                break;
            }
        }
        return false;
    }
    string getSum(string &num1, string &num2) {
        int val, flag = 0;
        int len1 = num1.length(), len2 = num2.length();
        string sum;
        while (len1 || len2) {
            val = 0;
            if (len1 > 0) {
                val += num1[len1 - 1] - '0';
                --len1;
            }
            if (len2 > 0) {
                val += num2[len2 - 1] - '0';
                --len2;
            }
            sum += to_string((val + flag) % 10);
            flag = (val + flag) / 10;
        }
        if (flag) {
            sum += '1';
        }
        reverse(sum.begin(), sum.end());
        return sum;
    }
};