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Additive Number (M)

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

Follow up: How would you handle overflow for very large input integers?

题意

给定一个字符串，判断能否将该字符串切割成若干个数字，且每个数字都是前两个数字之和。

思路

先确定好最开始的两个数，再递归判断整个字符串。

代码实现

Java

class Solution {
    public boolean isAdditiveNumber(String num) {
        for (int i = 1; i <= num.length() / 2; i++) {
            if (i > 1 && num.charAt(0) == ‘0‘) break;
            for (int j = 1; j <= num.length() / 2 && i + j < num.length(); j++) {
                if (j > 1 && num.charAt(i) == ‘0‘) break;
                String num1 = num.substring(0, i), num2 = num.substring(i, i + j);
                if (dfs(num.substring(i + j), add(num1, num2), num2)) {
                    return true;
                }
            }
        }
        return false;
    }

    private boolean dfs(String s, String target, String pre) {
        if (s.isEmpty()) {
            return true;
        }

        if (!s.startsWith(target)) {
            return false;
        }

        return dfs(s.substring(target.length()), add(target, pre), target);
    }

    private String add(String s, String t) {
        String sum = "";
        int i = 1;
        int carry = 0;
        while (i <= s.length() || i <= t.length()) {
            int num1 = 0, num2 = 0;
            if (i <= s.length()) {
                num1 = s.charAt(s.length() - i) - ‘0‘;
            }
            if (i <= t.length()) {
                num2 = t.charAt(t.length() - i) - ‘0‘;
            }
            int tmp = num1 + num2 + carry;
            carry = tmp / 10;
            sum = tmp % 10 + sum;
            i++;
        }
        if (carry != 0) {
            sum = carry + sum;
        }
        return sum;
    }
}

0306. Additive Number (M)

标签：targe   sequence   开始   ann   lse   const   over   number   思路   

原文地址：https://www.cnblogs.com/mapoos/p/13258690.html