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Question
Formatted question description: https://leetcode.ca/all/306.html
An additive number is a string whose digits can form an 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, return true
if it is an additive number or false
otherwise.
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:
1 <= num.length <= 35
num
consists only of digits.
Follow up: How would you handle overflow for very large input integers?
Algorithm
Let the first digit start with one digit, and the second digit will be searched for higher digits. The first two
digits are confirmed, and the third digit is obtained by adding them.
The three arrays are arranged to form a string, which is the same length as the original string if it is less than the original length, then take out the second and third numbers from the previous calculation, and use the same method to get the third number, then add the current string, and then sum Compared to the original string length.
Code
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import java.util.ArrayList; public class Additive_Number { public static void main(String[] args) { Additive_Number out = new Additive_Number(); Solution s = out.new Solution(); System.out.println(s.isAdditiveNumber("112358")); } // very good follow up analysis class Solution { boolean isFound = false; public boolean isAdditiveNumber(String num) { if (num == null || num.length() == 0) { return false; } dfs(num, 0, new ArrayList<Long>()); return isFound; } private void dfs(String num, int startIndex, ArrayList<Long> out) { if (isFound) { return; } if (startIndex == num.length()) { // @note: input is [1,2], then should return false // @note: not working if ==3, must be >=3 // input: "112358", out will be "[1, 1, 2, 3, 5, 8]" if (out.size() >= 3) { isFound = true; } return; } for (int i = startIndex; i < num.length(); i++) { String current = num.substring(startIndex, i + 1); // @note:+1 // skip invalid // 检测str的长度大于19就break,因为long的十进制数长度是19位 if ((current.length() > 1 && current.charAt(0) == '0') || current.length() > 19) { break; // because all the rest will all start with 0 } long curNum = Long.valueOf(current); long n = (long) out.size(); if (out.size() >= 2 && curNum != out.get((int) (n - 1)) + out.get((int) (n - 2))) { continue; } out.add(curNum); dfs(num, i + 1, out); out.remove(out.size() - 1); } } } } ############ class Solution { public boolean isAdditiveNumber(String num) { int n = num.length(); for (int i = 1; i < Math.min(n - 1, 19); ++i) { for (int j = i + 1; j < Math.min(n, i + 19); ++j) { if (i > 1 && num.charAt(0) == '0') { break; } if (j - i > 1 && num.charAt(i) == '0') { continue; } long a = Long.parseLong(num.substring(0, i)); long b = Long.parseLong(num.substring(i, j)); if (dfs(a, b, num.substring(j))) { return true; } } } return false; } private boolean dfs(long a, long b, String num) { if ("".equals(num)) { return true; } if (a + b > 0 && num.charAt(0) == '0') { return false; } for (int i = 1; i < Math.min(num.length() + 1, 19); ++i) { if (a + b == Long.parseLong(num.substring(0, i))) { if (dfs(b, a + b, num.substring(i))) { return true; } } } return false; } }
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class Solution: def isAdditiveNumber(self, num: str) -> bool: def dfs(startIndex: int, out: List[int]) -> bool: # equal check is here if len(out) >= 3 and out[-1] != out[-2] + out[-3]: return False if startIndex == len(num): return len(out) >= 3 for i in range(startIndex, len(num)): current = num[startIndex: i + 1] if (len(current) > 1 and current[0] == '0'): break out.append(int(current)) if dfs(i + 1, out): return True out.pop() return False if not num: return False return dfs(0, []) class Solution: def isAdditiveNumber(self, num: str) -> bool: def dfs(a, b, num): if not num: return True if a + b > 0 and num[0] == '0': return False for i in range(1, len(num) + 1): if a + b == int(num[:i]): if dfs(b, a + b, num[i:]): return True return False n = len(num) for i in range(1, n - 1): # 1st cut if i > 1 and num[0] == '0': # 0 + 1 = 1 is fine, but 00 + 1 is wrong break for j in range(i + 1, n): # 2nd cut, so making it a 3-segments if j - i > 1 and num[i] == '0': break if dfs(int(num[:i]), int(num[i:j]), num[j:]): # better than below, early stop return True return False ######## class Solution(object): def isAdditiveNumber(self, num): """ :type num: str :rtype: bool """ n = len(num) for x in range(0, n / 2): if x > 0 and num[0] == "0": break for y in range(x + 1, n): if y - x > 1 and num[x + 1] == "0": break i, j, k = 0, x, y while k < n: a = int(num[i:j + 1]) b = int(num[j + 1:k + 1]) add = str(int(a + b)) if not num.startswith(add, k + 1): break if len(add) + 1 + k == len(num): return True i = j + 1 j = k k = k + len(add) return False
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class Solution { public: bool isAdditiveNumber(string num) { int n = num.size(); for (int i = 1; i < min(n - 1, 19); ++i) { for (int j = i + 1; j < min(n, i + 19); ++j) { if (i > 1 && num[0] == '0') break; if (j - i > 1 && num[i] == '0') continue; auto a = stoll(num.substr(0, i)); auto b = stoll(num.substr(i, j - i)); if (dfs(a, b, num.substr(j, n - j))) return true; } } return false; } bool dfs(long long a, long long b, string num) { if (num == "") return true; if (a + b > 0 && num[0] == '0') return false; for (int i = 1; i < min((int) num.size() + 1, 19); ++i) if (a + b == stoll(num.substr(0, i))) if (dfs(b, a + b, num.substr(i, num.size() - i))) return true; return false; } };
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func isAdditiveNumber(num string) bool { n := len(num) var dfs func(a, b int64, num string) bool dfs = func(a, b int64, num string) bool { if num == "" { return true } if a+b > 0 && num[0] == '0' { return false } for i := 1; i < min(len(num)+1, 19); i++ { c, _ := strconv.ParseInt(num[:i], 10, 64) if a+b == c { if dfs(b, c, num[i:]) { return true } } } return false } for i := 1; i < min(n-1, 19); i++ { for j := i + 1; j < min(n, i+19); j++ { if i > 1 && num[0] == '0' { break } if j-i > 1 && num[i] == '0' { continue } a, _ := strconv.ParseInt(num[:i], 10, 64) b, _ := strconv.ParseInt(num[i:j], 10, 64) if dfs(a, b, num[j:]) { return true } } } return false } func min(a, b int) int { if a < b { return a } return b }