LeetCode 306. Additive Number

原题链接在这里：https://leetcode.com/problems/additive-number/

题目：

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

Constraints:

num consists only of digits '0'-'9'.
1 <= num.length <= 35

题解：

Get the first two numbers. And check if their sum could be got from the rest.

DFS state needs first number, second number and rest string.

If first 2 numbers are legal, calculate the sum.

If sum is equal to rest, return true.

Otherwise, if sum could be get from rest string. Continue DFS with 2nd, sum and new rest.

When getting the first 2 numbers, neither of them length could be larger than half of num length.

Time Complexity: exponential.

Space: O(n). n =sum.length(). stack space.

AC Java:

 1 class Solution {
 2     public boolean isAdditiveNumber(String num) {
 3         if(num == null || num.length() < 3){
 4             return false;
 5         }
 6         
 7         for(int i = 1; i<=num.length()/2; i++){
 8             for(int j = 1; j<=num.length()/2; j++){
 9                 if(dfs(num.substring(0, i), num.substring(i, i+j), num.substring(i+j))){
10                     return true;
11                 }
12             }
13         }
14         
15         return false;
16     }
17     
18     private boolean dfs(String a, String b, String c){
19         if((a.length() > 1 && a.charAt(0) == '0') || (b.length() > 1 && b.charAt(0) == '0')){
20             return false;
21         }
22         
23         String sum = getSum(a, b);
24         if(sum.equals(c)){
25             return true;
26         }
27         
28         if(sum.length() >= c.length() || !c.substring(0, sum.length()).equals(sum)){
29             return false;
30         }
31         
32         return dfs(b, c.substring(0, sum.length()), c.substring(sum.length()));
33     }
34     
35     private String getSum(String a, String b){
36         StringBuilder sb = new StringBuilder();
37         int i = a.length()-1;
38         int j = b.length()-1;
39         int carry = 0;
40         while(i>=0 || j>=0 || carry>0){
41             int temp = (i>=0 ? a.charAt(i)-'0' : 0) + (j>=0 ? b.charAt(j)-'0' : 0) + carry;
42             sb.insert(0, temp%10);
43             carry = temp/10;
44             i--;
45             j--;
46         }
47         
48         return sb.toString();
49     }
50 }