# Question

Formatted question description: https://leetcode.ca/all/306.html

 306	Additive Number

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



# Algorithm

Let the first digit start with one digit, and the second digit will be searched for higher digits. The first two digits are confirmed, and the third digit is obtained by adding them.

The three arrays are arranged to form a string, which is the same length as the original string if it is less than the original length, then take out the second and third numbers from the previous calculation, and use the same method to get the third number, then add the current string, and then sum Compared to the original string length.

# Code

Java

• import java.util.ArrayList;

public static void main(String[] args) {
Solution s = out.new Solution();

}

// very good follow up analysis
class Solution {

boolean isFound = false;

if (num == null || num.length() == 0) {
return false;
}

dfs(num, 0, new ArrayList<Long>());

return isFound;
}

private void dfs(String num, int startIndex, ArrayList<Long> out) {

if (isFound) {
return;
}

if (startIndex == num.length()) {
// @note: input is [1,2], then should return false
// @note: not working if ==3, must be >=3
//          input: "112358", out will be "[1, 1, 2, 3, 5, 8]"
if (out.size() >= 3) {
isFound = true;
}
return;
}

for (int i = startIndex; i < num.length(); i++) {

String current = num.substring(startIndex, i + 1); // @note：+1

// skip invalid
// 检测str的长度大于19就break，因为long的十进制数长度是19位
if ((current.length() > 1 && current.charAt(0) == '0') || current.length() > 19) {
}

long curNum = Long.valueOf(current);
long n = (long) out.size();

if (out.size() >= 2 && curNum != out.get((int) (n - 1)) + out.get((int) (n - 2))) {
continue;
}

dfs(num, i + 1, out);
out.remove(out.size() - 1);
}
}
}
}

• Todo

• class Solution(object):
"""
:type num: str
:rtype: bool
"""
n = len(num)
for x in range(0, n / 2):
if x > 0 and num[0] == "0":
break
for y in range(x + 1, n):
if y - x > 1 and num[x + 1] == "0":
break
i, j, k = 0, x, y
while k < n:
a = int(num[i:j + 1])
b = int(num[j + 1:k + 1])
if not num.startswith(add, k + 1):
break
if len(add) + 1 + k == len(num):
return True
i = j + 1
j = k