问题描述:
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.
解决思路:
首先对每一个运算符号进行分离左右两部分,若其中一部分是表达式,则继续遍历分离直至左右两边都是一个数字/符号,接着则通过运算符运算叠加。
具体代码实现:
vector<int> diffWaysToCompute(string input) {
vector<int> res;
int size = input.size();
for (int i = 0; i < size; i++) {
char c = input[i];
if (c == '+' || c == '-' || c == '*') {
vector<int> left = diffWaysToCompute(input.substr(0, i));
vector<int> right = diffWaysToCompute(input.substr(i + 1));
for (int j = 0; j < left.size(); j++) {
for (int k = 0; k < right.size(); k++) {
if (c == '+') {
res.push_back(left[j] + right[k]);
}
else if (c == '-') {
res.push_back(left[j] - right[k]);
}
else if (c == '*') {
res.push_back(left[j] * right[k]);
}
}
}
}
}
if (res.empty()) {
res.push_back(atoi(input.c_str()));
}
return res;
}