[LeetCode] Additive Number

yibinnn 发布于2019-08-14 16:38 / 3107人阅读

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?

Note

Here is my thought:
you get two consecutive substrings with two for loops, then call a recursive function to check whether the rest string starts with the sum of them.

Solution

public class Solution {
    public boolean isAdditiveNumber(String num) {
        int n = num.length();
        for (int i = 1; i <= (n-1)/2; i++) {
            if (num.charAt(0) == "0" && i > 1) break;
            for (int j = i+1; j-i <= n-j && i <= n-j; j++) {
                if (num.charAt(i) == "0" && j > i+1) break;
                long num1 = Long.parseLong(num.substring(0,i));
                long num2 = Long.parseLong(num.substring(i,j));
                String substr = num.substring(j);
                if (isValid(num1, num2, substr)) return true;
            }
        }
        return false;
    }
    public boolean isValid(long num1, long num2, String str) {
        if (str.equals("")) return true;
        long sum = num1+num2;
        String s = ((Long)sum).toString();
        if (!str.startsWith(s)) return false;
        return isValid(num2, sum, str.substring(s.length()));
    }
}

GPU云服务器云服务器 Additive LeetCode LeetCode_String leetcode_35_SearchIn

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