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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 *.
Example 1
Input: “2-1-1”.
((2-1)-1) = 0
(2-(1-1)) = 2
Output: [0, 2]
Example 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.
这题不是很好想,通过率可不低啊。
解法1:回溯法列出所有括号组合,分别用栈来计算。
有点复杂。这可以当做两个题目来做了。
解法2:和 Unique Binary Search Trees II 类似。分治的思想。
难点是要想到用符号当做分界点。
用递归实现。非递归还没想好。
class Solution {
public:
vector<int> diffWaysToCompute(string input) {
vector<int> res;
int data=0;
int i;
for(i=0;i<input.length()&&isdigit(input[i]);i++){
data=data*10+input[i]-'0';
}
if(i==input.length()){
res.push_back(data);
return res;
}
vector<int> left,right;
for(int i=0;i<input.length();i++){
if(isdigit(input[i])) continue;
left=diffWaysToCompute(input.substr(0,i));
right=diffWaysToCompute(input.substr(i+1,input.length()-1-i));
for(int l=0;l<left.size();l++){
for(int r=0;r<right.size();r++){
res.push_back(calculate(left[l],right[r],input[i]));
}
}
}
return res;
}
private:
int calculate(int a, int b, char op) {
switch (op) {
case '+': return a + b;
case '-': return a - b;
case '*': return a * b;
}
return 1;
}
};