306. Additive Number

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.

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?

class Solution public boolean isAdditiveNumber(String num) { int L = num.length(); //确定第一个数，最终用num.subStr(0,i)来确定第一个数，所以i可以用来标示第一个数的长度， //但是下标i不包含在第一个数中 for(int i=1;i<=(L-1)/2;i++){ //如果长度大于等于2，则不能以0开头 if(num.startsWith("0") && i>=2) break; //确定第二个数，第一个数用num.subStr(i,j),包括i，不包括j，所以长度为j-i, //第三个数从下标j开始，长度最长为L-1-j+1，即L-j for(int j=i+1;(L-j)>=i && (L-j)>=j-i;j++){ if(num.charAt(i)=='0' && j-i>=2) break; long num1 = Long.parseLong(num.substring(0,i)); long num2 = Long.parseLong(num.substring(i,j)); if(isAdditive(num.substring(j), num1, num2)){ return true; } } } return false; } //判断由num1,num2和后续的字串能否构成加法序列 public boolean isAdditive(String remain,long num1,long num2){ if(remain.equals("")) return true; long sum = num1+num2; String sumStr = ""+sum; if(!remain.startsWith(sumStr)) return false; return isAdditive(remain.substring(sumStr.length()), num2, sum); }}

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