标签:led Plan 一半 war 字符 orm lead substring 大于
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
Follow up:
How would you handle overflow for very large input integers?
class Solution { public boolean isAdditiveNumber(String num) { int k = num.length(); if(k < 3) return false; for(int i = 1; i <= num.length() / 2; i++){ //选第一个数字A,长度从1到最大长度的一半(11011,A=11) if(num.charAt(0) == ‘0‘ && i >= 2) return false;//如果第一个字符是0,而且A大于1位数就返回false Long a = Long.valueOf(num.substring(0, i)); for(int j = 1; k - i - j >= Math.max(i, j); j++){ if(num.charAt(i) == ‘0‘ && j > 1) break; Long b = Long.valueOf(num.substring(i, i+j)); if(helper(a, b, i+j, num)) return true; } } return false; } public boolean helper(Long a, Long b, int begin, String num){ if(begin == num.length()) return true; b = a + b; a = b - a; String res = b.toString(); return num.startsWith(res, begin) && helper(a, b, begin + res.length(), num); } }
https://www.youtube.com/watch?v=LziJZT2uRwc
不错,细节挺多的
标签:led Plan 一半 war 字符 orm lead substring 大于
原文地址:https://www.cnblogs.com/wentiliangkaihua/p/11548777.html