LeetCode 241. Different Ways to Add Parentheses

Description

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]

Solution

运用分治思想，遇到运算符时把问题分为两个子问题，并对两个子问题的结果求解，超时

vector<int> diffWaysToCompute(string input) {
vector<int> result;
for (int i = 0; i < input.size(); i++) {
char c = input[i];
if (ispunct(c)) {
vector<int> leftResult = diffWaysToCompute(input.substr(0, i));
vector<int> rightResult = diffWaysToCompute(input.substr(i + 1));
for (int j = 0; j < leftResult.size(); j++) {
for (int k = 0; k < rightResult.size(); k++) {
switch (c) {
case '+':
result.push_back(leftResult[j] + rightResult[k]);
break;
case '-':
result.push_back(leftResult[j] - rightResult[k]);
break;
case '*':
result.push_back(leftResult[j] * rightResult[k]);
break;
default:
break;
}
}
}
}
}
if (result.empty()) {
result.push_back(stoi(input));
return result;
}
}

注意

if (ispunct(c))

，如果参数是除字母，数字和空格外可打印字符，函数返回非零值，否则返回零值。此处表示if(c == ‘+’ || c == ‘-’ || c == ‘*’)

显然上面这种方法会有很多对子问题的重复计算，效率不高。这里使用一个哈希表来保存已经计算过子问题的结果，下次需要用时直接取出，如果没有则进行计算。

#include <iostream>
#include <vector>
#include <map>
using namespace std;

vector<int> diffWaysToCompute(string input, map<string, vector<int>>&map1) {
vector<int> result;
for (int i = 0; i < input.size(); i++){
char c = input[i];
if (ispunct(c)){
string leftStr = input.substr(0, i);
string rightStr = input.substr(i + 1);
vector<int> leftResult;
vector<int> rightResult;

if(map1.find(leftStr) != map1.end())
leftResult = map1[leftStr];
else
leftResult = diffWaysToCompute(leftStr, map1);
if(map1.find(rightStr) != map1.end())
rightResult = map1[rightStr];
else
rightResult = diffWaysToCompute(rightStr, map1);

for (int j = 0; j < leftResult.size(); j++) {
for (int k = 0; k < rightResult.size(); k++) {
switch (c) {
case '+':
result.push_back(leftResult[j] + rightResult[k]);
break;
case '-':
result.push_back(leftResult[j] - rightResult[k]);
break;
case '*':
result.push_back(leftResult[j] * rightResult[k]);
break;
default:
break;
}
}
}
}
}
if (result.empty())
result.push_back(stoi(input));
map1[input] = result;
return result;

}

int main(){
map<string, vector<int>> map1;
string input = "2*3-4*5";
vector<int> vec;
vec = diffWaysToCompute(input, map1);
for(int i = 0; i < vec.size(); i++)
cout<<vec[i]<<' ';
cout<<endl;