LeetCode 306 Additive Number

LeetCode

题目描述

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?

分析

参考链接

变量 i 和 j 分别是第一个数和第二个数的长度，然后以此检验这个序列是否为 Additive Number。两个数的和 sum 的长度要大于 i 和 j。

考虑溢出问题，可以使用 BigInteger；不考虑溢出问题，可以使用 Long。

代码

package LeetCode;
public class L306_Additive_Number {
    public boolean isAdditiveNumber(String num) {
        int n = num.length();
        // i是第一个数的长度，j是第二个数的长度
        for (int i = 1; i <= n / 2; i++) {
            for (int j = 1; Math.max(j, i) <= n - i - j; j++) {
                // 判断以长度i,j是否是Additive Number
                if (isValid(i, j, num)) {
                    return true;
                }
            }
        }
        return false;
    }
    private boolean isValid(int i, int j, String num) {
        if (num.charAt(0) == '0' && i > 1) {
            return false;
        }
        if (num.charAt(i) == '0' && j > 1) {
            return false;
        }
        String sum;
        Long x1 = Long.parseLong(num.substring(0, i));
        Long x2 = Long.parseLong(num.substring(i, i + j));
        for (int start = i + j; start < num.length(); start += sum.length()) {
            x2 = x1 + x2;
            x1 = x2 - x1;
            sum = x2.toString();
            if (!num.startsWith(sum, start)) {
                return false;
            }
        }
        return true;
    }
}