【LeetCode】147.Additive Number

最新推荐文章于 2025-03-22 19:42:57 发布

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最新推荐文章于 2025-03-22 19:42:57 发布

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本文探讨了一种算法，用于判断一个由数字字符组成的字符串是否为加法数。加法数是指其数字可以形成加法序列的字符串，序列中除前两个数外，每个后续数字都是前两个数字之和。文章提供了具体的实现代码，并通过实例展示了如何判断给定的字符串是否满足加法数的条件。

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题目描述(Medium)

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.

Given a string containing only digits '0'-'9', write a function to determine if it's an additive number.

Note: Numbers in the additive sequence cannot have leading zeros, so sequence 1, 2, 03 or 1, 02, 3 is invalid.

题目链接

https://leetcode.com/problems/additive-number /description/

Example 1:

Input: "112358"
Output: true
Explanation: 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

Example 2:

Input: "199100199"
Output: true
Explanation: The additive sequence is: 1, 99, 100, 199.
1 + 99 = 100, 99 + 100 = 199

Follow up:
How would you handle overflow for very large input integers?

算法分析

前两个数确定之后，后面数字的排列也就确定了，因此只需遍历前两个数字的组合，将生成序列与原始序列比较是否一致即可，由于考虑到大数加法的溢出，直接将字符串相加。

提交代码：

class Solution {
public:
    bool isAdditiveNumber(string num) {
        for (int i = 1; i < num.size(); ++i) {
            for (int j = 1; j < num.size() - i; ++j) {
                string c1 = num.substr(0, i);
                string c2 = num.substr(i, j);
                if ((c1.size() > 1 && c1[0] == '0') || 
                    (c2.size() > 1 && c2[0] == '0')) continue;
                string add = stringAdd(c1, c2);
                string now = c1 + c2 + add;
                
                while (now.size() < num.size()) {
                    c1 = c2;
                    c2 = add;
                    add = stringAdd(c1, c2);
                    now += add;
                }
                
                if (now == num) return true;
            }
        }
        
        return false;
    }

private:
    string stringAdd(string a, string b) {
        int carry = 0;
        int len1 = a.size(), len2 = b.size();
        if(len1 == 0) return b;
        else if(len2 == 0) return a;
        int length = max(len1, len2);
        string result(length, '0');
        char c1, c2;
        --len1, --len2, --length;
        for(; len1 >= 0 || len2 >= 0; --len1, --len2, --length) {
            if (len1 < 0) c1 = '0';
            else c1 = a[len1];            
            
            if (len2 < 0) c2 = '0';
            else c2 = b[len2];
            
            int res = c1 - '0' + c2 - '0' + carry;
            carry = res / 10;
            res %= 10;
            result[length] = '0' + res;
        }
        
        if (carry != 0) result = '1' + result;
        return result;
    }
    
};