306. Additive Number

 

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, 3is invalid.

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?

Credits:
Special thanks to @jeantimex for adding this problem and creating all test cases.

 

这道题定义了一种加法数,就是至少含有三个数字,除去前两个数外,每个数字都是前面两个数字的和,题目中给了许多例子,也限定了一些不合法的情况,比如两位数以上不能以0开头等等,让我们来判断一个数是否是加法数。目光犀利的童鞋应该一眼就能看出来,这尼玛不就是斐波那契数组么,跟另一道 Split Array into Fibonacci Sequence 简直不要太像啊。只不过那道题要返回一种组合方式,而这道题只是问能否拆成斐波那契数列。

开始我还想是否能用动态规划来解,可是发现不会写状态转移方程,只得作罢。其实这题可用Brute Force的思想来解,我们让第一个数字先从一位开始,第二个数字从一位,两位,往高位开始搜索,前两个数字确定了,相加得到第三位数字,三个数组排列起来形成一个字符串,和原字符串长度相比,如果小于原长度,那么取出上一次计算的第二个和第三个数,当做新一次计算的前两个数,用相同的方法得到第三个数,再加入当前字符串,再和原字符串长度相比,以此类推,直到当前字符串长度不小于原字符串长度,比较两者是否相同,相同返回true,不相同则继续循环。如果所有情况都遍历完了还是没有返回true,则说明不是Additive Number,返回false,参见代码如下:

 

解法一:

class Solution {
public:
    bool isAdditiveNumber(string num) {
        for (int i = 1; i < num.size(); ++i) {
            string s1 = num.substr(0, i);
            if (s1.size() > 1 && s1[0] == '0') break;
            for (int j = i + 1; j < num.size(); ++j) {
                string s2 = num.substr(i, j - i);
                long d1 = stol(s1), d2 = stol(s2);
                if ((s2.size() > 1 && s2[0] == '0')) break;
                long next = d1 + d2;
                string nextStr = to_string(next);
                if (nextStr != num.substr(j, nextStr.length())) continue; // optimization here
                string allStr = s1 + s2 + nextStr;
                while (allStr.size() < num.size()) {
                    d1 = d2;
                    d2 = next;
                    next = d1 + d2;
                    nextStr = to_string(next);
                    allStr += nextStr;
                }
                if (allStr == num) return true;
            }
        }
        return false;
    }
};

 

此题还有递归解法,博主最先尝试的是跟那道 Split Array into Fibonacci Sequence 一样的递归方法,虽然这道题不用返回组合方式,但是我们仍可以使用一样的递归来做,只不过这里的结果res是一个布尔型的全局变量,当我们找到一组符合题意的组合了之后,就将结果res设置为true,这样一旦再进入递归函数时,只要res为true了,就直接返回即可。之后的部分就基本相同了,可以参见 Split Array into Fibonacci Sequence  中的讲解,注意稍有不同的地方是,这里拆分出来的数字是可以超过整型int范围的,但是貌似不会超过长整型long的范围,所以我们可以加一个检测str的长度大于19就break,因为long的十进制数长度是19位,参见代码如下:

 

解法二:

class Solution {
public:
    bool isAdditiveNumber(string num) {
        bool res = false;
        vector<long> out;
        helper(num, 0, out, res);
        return res;
    }
    void helper(string& num, int start, vector<long>& out, bool& res) {
        if (res) return;
        if (start >= num.size()) {
            if (out.size() >= 3) res = true; 
            return;
        }
        for (int i = start; i < num.size(); ++i) {
            string str = num.substr(start, i - start + 1);
            if ((str.size() > 1 && str[0] == '0') || str.size() > 19) break;
            long curNum = stol(str), n = out.size();
            if (out.size() >= 2 && curNum != out[n - 1] + out[n - 2]) continue;
            out.push_back(curNum);
            helper(num, i + 1, out, res);
            out.pop_back();
        }
    }
};

 

由于这道题并不需要我们返回具体的组合方式,所以也可以不使用上面的写法,而是不停的用当前的两个数,去拼后面的数字,一旦拼不出来了,直接返回false。跟解法一类似,首先用两个for循环来确定前两个数字,然后将后面的整体提取出来,和当前的两个数字一起调用递归,若递归返回true了,则直接返回true,否则继续遍历。

在递归函数中,还是首先检验是否存在leading zeros,然后将 num1 和 num2 分别转为长整型long,相加后再转回字符串,存入sumStr,这时候看sumStr是否和num相等,是的话直接返回true。否则再来检验,若sumStr的长度大于等于num了,返回false,或者在num中取跟sumStr长度相等的子串,若不等于sumStr,说明无法拼出来,也返回false。若都没返回的话,就再次调用递归,不过这次num要去掉和sumStr长度相等的子串,留下后面的部分,此时带入递归的两个数字要变成 num2 和 sumStr,继续跟后面的比较,参见代码如下:

 

解法三:

class Solution {
public:
    bool isAdditiveNumber(string num) {
        for (int i = 1; i < num.size(); ++i) {
            string s1 = num.substr(0, i);
            if (s1.size() > 1 && s1[0] == 0) break;
            for (int j = i + 1; j < num.size(); ++j){
                string s2 = num.substr(i, j - i);
                if (s2.size() > 1 && s2[0] == 0) break;
                if(helper(num.substr(j), s1, s2)) return true;
            }
        }
        return false;
    }
    bool helper(string num, string num1, string num2){
        if ((num1.size() > 1 && num1[0] == '0') || (num2.size() > 1 && num2[0] == '0')) return false;
        string sumStr = to_string(stol(num1) + stol(num2));
        if (sumStr == num) return true;
        if (sumStr.size() >= num.size() || num.substr(0, sumStr.size()) != sumStr) return false;
        return helper(num.substr(sumStr.size()), num2, sumStr);
    }
};

 

题目中有个 follow up 说是让我们handle超大整数的溢出情况,那么我们想,上面的解法中哪块可能溢出啊,当然只有将子串转为长整型long的时候,假如此时字符串特别长,甚至超出了长整型的范围,那么我们就不能用stol了,此时就不能转换了,那么我们只能强行将两个字符串相加了,这里用的又是另外一道题目 Add Strings 的知识点了,这样我们就完美的避开了可能溢出的情况了,参见代码如下:

 

解法四:

// Follow up, handle overflow for very large input integers.
class Solution {
public:
    bool isAdditiveNumber(string num) {
        for (int i = 1; i < num.size(); ++i) {
            string s1 = num.substr(0, i);
            if (s1.size() > 1 && s1[0] == 0) break;
            for (int j = i + 1; j < num.size(); ++j){
                string s2 = num.substr(i, j - i);
                if (s2.size() > 1 && s2[0] == 0) break;
                if(helper(num.substr(j), s1, s2)) return true;
            }
        }
        return false;
    }
    bool helper(string num, string num1, string num2){
        if ((num1.size() > 1 && num1[0] == '0') || (num2.size() > 1 && num2[0] == '0')) return false;
        string sumStr = add(num1, num2);
        if (sumStr == num) return true;
        if (sumStr.size() >= num.size() || num.substr(0, sumStr.size()) != sumStr) return false;
        return helper(num.substr(sumStr.size()), num2, sumStr);
    }
    string add(string num1, string num2) {
        string res;
        int i = (int)num1.size() - 1, j = (int)num2.size() - 1, carry = 0;
        while (i >= 0 || j >= 0) {
            int val1 = (i >= 0) ? (num1[i--] - '0') : 0;
            int val2 = (j >= 0) ? (num2[j--] - '0') : 0;
            int sum = val1 + val2 + carry;
            res.push_back(sum % 10 + '0');
            carry = sum / 10;
        }
        if (carry) res.push_back(carry + '0');
        reverse(res.begin(), res.end());
        return res;
    }
};

 

类似题目:

Add Strings 

Split Array into Fibonacci Sequence

 

参考资料:

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

https://leetcode.com/problems/additive-number/discuss/75567/Java-Recursive-and-Iterative-Solutions

https://leetcode.com/problems/additive-number/discuss/75704/My-Simple-C%2B%2B-Non-recursion-Solution

https://leetcode.com/problems/additive-number/discuss/75576/0ms-concise-C%2B%2B-solution-(perfectly-handles-the-follow-up-and-leading-0s)

 

LeetCode All in One 题目讲解汇总(持续更新中…)


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