96 lines
2.9 KiB
C++
96 lines
2.9 KiB
C++
// Source : https://leetcode.com/problems/additive-number/
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// Author : Hao Chen
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// Date : 2015-11-22
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/***************************************************************************************
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*
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* Additive number is a positive integer whose digits can form additive sequence.
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*
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* A valid additive sequence should contain at least three numbers. Except for the
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* first two numbers, each subsequent number in the sequence must be the sum of the
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* preceding two.
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*
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* For example:
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* "112358" is an additive number because the digits can form an additive sequence: 1,
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* 1, 2, 3, 5, 8.
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* 1 + 1 = 2, 1 + 2 = 3, 2 + 3 = 5, 3 + 5 = 8
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* "199100199" is also an additive number, the additive sequence is: 1, 99, 100, 199.
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* 1 + 99 = 100, 99 + 100 = 199
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*
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* Note: Numbers in the additive sequence cannot have leading zeros, so sequence 1, 2,
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* 03 or 1, 02, 3 is invalid.
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*
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* Given a string represents an integer, write a function to determine if it's an
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* additive number.
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*
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* Follow up:
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* How would you handle overflow for very large input integers?
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*
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* Credits:Special thanks to @jeantimex for adding this problem and creating all test
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* cases.
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*
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***************************************************************************************/
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class Solution {
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public:
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bool isAdditiveNumber(string num) {
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int len = num.size();
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for(int i=1; i<len/2+1; i++) {
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string n1 = num.substr(0, i);
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if ( n1.size()>1 && n1[0] == '0') break;
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for(int j=i+1; j<len; j++) {
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string n2 = num.substr(i, j-i);
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if ( n2.size()>1 && n2[0] == '0') break;
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string n3 = num.substr(j);
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if (isAdditiveNumberHelper(n1, n2, n3)) return true;
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}
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}
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return false;
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}
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private:
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bool isAdditiveNumberHelper(string& n1, string& n2, string& n3){
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string add = StringAdd(n1, n2);
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if (add.size() > n3.size()) return false;
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if (add == n3 ) return true;
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//split the n3 to 2 parts, and keep going.
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string cut = n3.substr(0, add.size());
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if (add == cut) {
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string rest = n3.substr(add.size());
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return isAdditiveNumberHelper(n2, add, rest);
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}
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return false;
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}
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string StringAdd(string n1, string n2) {
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if (n1.size() < n2.size()) {
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string tmp = n1;
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n1 = n2;
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n2 = tmp;
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}
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int carry=0;
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string result;
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for (int i=n1.size()-1, j=n2.size()-1; i>=0; i--, j--) {
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int n = n1[i] - '0' + carry;
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if ( j >= 0) {
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n += n2[j] - '0';
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}
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char ch = n % 10 + '0';
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carry = n/10;
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result = ch + result;
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}
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if (carry>0) result = (char)(carry+'0') + result;
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return result;
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}
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};
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