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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?
public class Solution { public boolean isAdditiveNumber(String num) { int l = num.length(); if(l<3) return false; if(num.startsWith("00")) //Then the whole string should be "000..." { for(int i=2;i<l;++i) { if(num.charAt(i) != ‘0‘) return false; } return true; } int firstNumEndIndex = (l-1)/2; if(num.startsWith("0")) //Then 0 must be the first interger. firstNumEndIndex = 1; for(int i = 1; i<= firstNumEndIndex; ++i) { String add1 = num.substring(0, i); //[0, i) => first number for(int j = i+1; j<= l - Math.max(i,j-i); ++j) { String add2 = num.substring(i, j);//[i, j) => second number if(add2.equals("0")) { if(rollToEnd(num, add1, add2, j)) return true; break; } if(rollToEnd(num, add1, add2, j)) return true; } } return false; } //[start,end) private boolean rollToEnd (String num, String add1, String add2, int from) { String sumString = getSumString(add1, add2); String rest = num.substring(from); if(rest.startsWith("0")) return false; if(rest.equals(sumString)) return true; if(rest.startsWith(sumString)) return rollToEnd(num, add2, sumString, from+sumString.length()); return false; } private String getSumString(String add1, String add2) { int extra = 0; int l1 = add1.length(); int l2 = add2.length(); Deque<Integer> sum = new ArrayDeque<Integer>(); while(l1>0 && l2>0) { int a1 = add1.charAt(l1-1) - ‘0‘; int a2 = add2.charAt(l2-1) - ‘0‘; int s = a1+a2+extra; extra = s/10; sum.push(s%10); --l1; --l2; } while(l1>0) { int a = add1.charAt(l1-1) - ‘0‘; int s = a + extra; extra = s/10; sum.push(s%10); --l1; } while(l2>0) { int a = add2.charAt(l2-1) - ‘0‘; int s = a + extra; extra = s/10; sum.push(s%10); --l2; } if(extra == 1) sum.push(extra); StringBuilder sb = new StringBuilder(); while(!sum.isEmpty()) sb.append(sum.pop()); return sb.toString(); } }
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原文地址:http://www.cnblogs.com/neweracoding/p/5597349.html