17
08/2015
[LeetCode] Different Ways to Add Parentheses
Different Ways to Add Parentheses
Given a string of numbers and operators, return all possible results from computing all the different possible ways to group numbers and operators. The valid operators are+, - and *.
Input: "2-1-1".
((2-1)-1) = 0 (2-(1-1)) = 2
Output: [0, 2]
Input: "2*3-4*5"
(2*(3-(4*5))) = -34 ((2*3)-(4*5)) = -14 ((2*(3-4))*5) = -10 (2*((3-4)*5)) = -10 (((2*3)-4)*5) = 10
Output: [-34, -14, -10, -10, 10]
Credits:
Special thanks to @mithmatt for adding this problem and creating all test cases.
解题思路；
这道题可以用递归的方法来做。对于一个输入字符串s，一次获得每个标点符号的左侧left和右侧right的值，然后两两组合成结果。
class Solution {
public:
vector<int> diffWaysToCompute(string input) {
vector<int> result;
int len = input.length();
if(len == 0){
return result;
}
if(isInt(input)){
result.push_back(std::stoi(input));
return result;
}
for(int i=0; i<len; i++){
if(input[i]<'0' || input[i]>'9'){
vector<int> leftResult = diffWaysToCompute(input.substr(0, i));
vector<int> rightResult = diffWaysToCompute(input.substr(i+1));
for(int m = 0; m<leftResult.size(); m++){
for(int n = 0; n<rightResult.size(); n++){
switch(input[i]){
case '+':
result.push_back(leftResult[m] + rightResult[n]);
break;
case '-':
result.push_back(leftResult[m] - rightResult[n]);
break;
case '*':
result.push_back(leftResult[m] * rightResult[n]);
break;
}
}
}
}
}
return result;
}
bool isInt(string& s){
int len = s.length();
for(int i=0; i<len; i++){
if(s[i]<'0' || s[i]>'9'){
return false;
}
}
return true;
}
};
0 条评论