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Description
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.
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?
分析
- 这道题直观的做法是暴力做法，非递归的方法如下
- 注意这里面用到了C++的一些字符串操作的函数，substr,stoll，to_string()等等，使得代码写得简洁易懂
代码
class Solution {
public:
bool isAdditiveNumber(string num) {
for(int i=1;i<num.size();i++){
for(int j=i+1;j<num.size();j++){
string s1=num.substr(0,i);
string s2=num.substr(i,j-i);
long long d1=stoll(s1.c_str());
long long d2=stoll(s2.c_str());
if((s1.size()>1&&s1[0]=='0')||(s2.size()>1&&s2[0]=='0')){
continue;
}
long long next=d1+d2;
string nexts = to_string(next);
string now=s1+s2+nexts;
while(now.size()<num.size()){
d1=d2;
d2=next;
next=d1+d2;
nexts=to_string(next);
now+=nexts;
}
if(now==num) return true;
}
}
return false;
}
};