给出一个直观的解法,将每个数写成 "+(nom)/(denom)" 的string形式放到deque里,然后再把deque两两计算合并。计算时,要用到gcd来消公约数。写的比较长。
class Solution {
public:
int gcd(int a, int b){
return b == 0 ? a : gcd(b, a % b);
}
string calculate(string s1, string s2){
int idx1 = s1.find_first_of('/'), idx2 = s2.find_first_of('/');
int nom_1 = stoi(s1.substr(0, idx1)), nom_2 = stoi(s2.substr(0, idx2));
int denom_1 = stoi(s1.substr(idx1+1)), denom_2 = stoi(s2.substr(idx2+1));
int res_denom = denom_1 * denom_2;
int res_nom = nom_1 * denom_2 + nom_2 * denom_1;
if(res_nom == 0) return"+0/1";
int common = gcd(abs(res_nom), abs(res_denom));
res_nom /= common; res_denom /= common;
int sign = (res_nom * res_denom < 0) ? -1 : 1;
return (sign == 1 ? '+' : '-') + to_string(abs(res_nom)) + '/' + to_string(abs(res_denom));
}
string fractionAddition(string expression) {
if(expression.empty()) return "";
if(isdigit(expression[0])) expression.insert(expression.begin(), '+');
deque<string> dq;
int start = 0;
for(int i=1; i<=expression.length(); i++){
if( i == expression.length() || expression[i] == '+' || expression[i] == '-'){
string temp = expression.substr(start, i-start);
dq.push_back(temp);
start = i;
}
}
while(dq.size() >= 2){
string s1 = dq.front(); dq.pop_front();
string s2 = dq.front(); dq.pop_front();
string res = calculate(s1, s2);
//cout << s1 << " " << s2 << " " << res << endl;
dq.push_front(res);
}
string res = dq.front();
return res[0] == '-' ? res : res.substr(1);
}
};
更简洁的solution参见如下,用regular expression split,再loop数组相加.
https://discuss.leetcode.com/topic/89991/concise-java-solution
其中,String[] fracs = expression.split("(?=[-+])"); 运用了zero-positive look ahead, 表明在+或-的前面split.
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