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  • Leetcode: 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?

    The method is to use backtracking with pruning,

    we handle the case of overflow when using very large input integers by using long

    感想:第1,2个数很重要,决定了后面所有的数,所以先把第1,2个数确定出来,再backtracking看能不能走到最后,第1个数起点固定,所以通过定义第1,2个数终点来确定第1,2个数

     1 public class Solution {
     2     public boolean isAdditiveNumber(String num) {
     3         if (num==null || num.length()<3) return false;
     4         int len = num.length();
     5         for (int i=1; i<=len-2; i++) {
     6             if (i>1 && num.charAt(0)=='0') break;
     7             for (int j=i+1; j<=len-1; j++) {
     8                 if (j>i+1 && num.charAt(i)=='0') break;
     9                 int start1=0, start2=i, start3=j;
    10                 while (start3 < len) {
    11                     long first = Long.parseLong(num.substring(start1, start2));
    12                     long second = Long.parseLong(num.substring(start2, start3));
    13                     long third = first + second;
    14                     if (num.substring(start3).startsWith(String.valueOf(third))) {
    15                         start1 = start2;
    16                         start2 = start3;
    17                         start3 = start3 + String.valueOf(third).length();
    18                     }
    19                     else break;
    20                 }
    21                 if (start3 == len) return true;
    22             }
    23         }
    24         return false;
    25     }
    26 }
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  • 原文地址:https://www.cnblogs.com/EdwardLiu/p/5087609.html
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