@Yano 2016-03-22T02:50:49.000000Z 字数 1430 阅读 1900

# 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.

How would you handle overflow for very large input integers?

# 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;    }}

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