Leetcode: 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

*

.

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]

类似于求所有的二叉树，乘法法则。考虑了负数的情况，简单递归实现。可以保存中间结果来优化一下：查表，如果以前计算过，则不需要重新计算。

class Solution {
public:
vector<int> diffWaysToCompute(string input) {
vector<int> result;
bool negative = false;
if (!input.empty() && input[0] == '-') {
negative = true;
input = input.substr(1);
}

for (int i = 0; i < input.size(); ++i) {
if (input[i] != '+' && input[i] != '-' && input[i] != '*') {
continue;
}

string leftPart = negative ? ("-" + input.substr(0, i)) : input.substr(0, i);
vector<int> left = diffWaysToCompute(leftPart);
vector<int> right = diffWaysToCompute(input.substr(i + 1));
for (int ileft = 0; ileft < left.size(); ++ileft) {
for (int iright = 0; iright < right.size(); ++iright) {
if (input[i] == '+') {
result.push_back(left[ileft] + right[iright]);
}
else if (input[i] == '-') {
result.push_back(left[ileft] - right[iright]);
}
else {
result.push_back(left[ileft] * right[iright]);
}
}
}
}

if (result.empty()) {
int val = atoi(input.c_str());
if (negative) {
val = -val;
}
result.push_back(val);
}

return result;
}
};