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?
题目大意:
判断字符串是否可以被划分成一个一个字符串字串,满足加法性质。
例如:"112358",可以被划分为 1, 1, 2, 3, 5, 8. 满足1 + 1 = 2, 1 + 2 = 3, 2 + 3 = 5, 3 + 5 = 8条件。
解法:
首先如果一个字符串满足这样的加法性质,那么只要确定前两个数字,那么后面的数字都会自动被确定,所以只需要确定前两个数字,就可以得到这个字符串的这个划分是否满足要求。题目要求有大数要求,所以使用BigInteger类。
import java.math.BigInteger;
class Solution {
private boolean isVaild(BigInteger x1,BigInteger x2,int start,String num){
if (start==num.length()) return true;
x2=x1.add(x2);
x1=x2.subtract(x1);
String res=x2.toString();
return num.startsWith(res,start)&&isVaild(x1,x2,start+res.length(),num);
}
public boolean isAdditiveNumber(String num) {
int n=num.length();
for (int i=1;i<=num.length()/2;i++){
if (num.charAt(0)==‘0‘&&i>1) return false;
BigInteger x1=new BigInteger(num.substring(0,i));
for (int j=1;Math.max(i,j)<=n-i-j;j++){
if (num.charAt(i)==‘0‘&&j>1) break;
BigInteger x2=new BigInteger(num.substring(i,i+j));
if (isVaild(x1,x2,i+j,num)){
return true;
}
}
}
return false;
}
}
原文:https://www.cnblogs.com/xiaobaituyun/p/10886498.html