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.
public class Solution { public boolean isAdditiveNumber(String num) { if(num == null || num.length() == 0) return false; int n = num.length(); for(int i = 1; i <= n/2; i++){ // len of x1 for(int j = 1; Math.max(i, j) + i + j <= n; j++) { // len of x2 if(isValid(i, j, num)) return true; } } return false; } public boolean isValid(int i, int j, String num){ if(i > 1 && num.charAt(0) == "0") return false; if(j > 1 && num.charAt(i) == "0") 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; // sum of x1 and x2, and became new x2 x1 = x2 - x1; // last x2 is new x1 sum = x2.toString(); if(!num.startsWith(sum, start)) return false; } return true; } }
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摘要:為了減少無效遍歷,我們可以在尋找第一個數字和第二個數字的時候及時終止。我們可以知道第一個數字的長度不應該超過字符串長度的一般,第二個數字的長度無法超過字符串長度減去第一個數字的長度。因此一旦遇到,在判斷完作為加數時是否合法后,直接跳出循環。 題目要求 Additive number is a string whose digits can form additive sequence....
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摘要:題目解答不越界長度的當可以走到后面沒有和了的時候,說明這個滿足條件直接可以知道這個是不是存在于中越界長度的越界長的度 題目:Additive number is a string whose digits can form additive sequence. A valid additive sequence should contain at least three numbers...
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 ...
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