<LeetCode OJ> 241. Different Ways to Add Parentheses

Total Accepted: 23674 Total
Submissions: 65569 Difficulty: Medium

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.

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Divide and Conquer

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分析：
本题不会，代码纯属借鉴和学习！

这是一个要考虑各种组合情况的问题，关键还是怎么入手得到各种组合情况！

从别人的代码来看，一旦遇到运算符号(不是运算符号暂时不处理)，就分治法考虑各种组合情况：

然后将运算结果重新根据当前运算符进行运算。

class Solution {
public:
vector<int> diffWaysToCompute(string input) {
vector<int> result;
int size = input.size();
for (int i = 0; i < size; i++) {
char cur = input[i];
if (cur == '+' || cur == '-' || cur == '*') {//如果是运算符号
// 分治法，分两边解决，解决获取结果之后重新根据当前符号进行运算
vector<int> result1 = diffWaysToCompute(input.substr(0, i));//从0位置开始获取长度为i的字符串
vector<int> result2 = diffWaysToCompute(input.substr(i+1));//从0位置开始获取后面所有字符
for (auto n1 : result1) {
for (auto n2 : result2) {
if (cur == '+')
result.push_back(n1 + n2);
else if (cur == '-')
result.push_back(n1 - n2);
else
result.push_back(n1 * n2);
}
}
}
}
// 如果输入字符串只是数字（字符数字）
if (result.empty())
result.push_back(atoi(input.c_str()));
return result;
}
};

完整代码：

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

vector<int> diffWaysToCompute(string input) {
vector<int> result;
int len=input.size();
for(int k=0;k<len;k++)
{
if(input[k]=='+'||input[k]=='-'||input[k]=='*')
{
vector<int> result1=diffWaysToCompute(input.substr(0,k));
vector<int> result2=diffWaysToCompute(input.substr(k+1));
for(vector<int>::iterator i=result1.begin();i!=result1.end();i++)
for(vector<int>::iterator j=result2.begin();j!=result2.end();j++)
{
if(input[k]=='+')
result.push_back((*i)+(*j));
else if(input[k]=='-')
result.push_back((*i)-(*j));
else
result.push_back((*i)*(*j));
}
}
}
if(result.empty())
result.push_back(atoi(input.c_str()));
return result;
}
int main()
{
string input="2*3-4*5";
vector<int> vec;
vec=diffWaysToCompute(input);
for(int i=0;i<vec.size();i++)
cout<<vec[i]<<' ';
cout<<endl;
system("pause");
}

注：本博文为EbowTang原创，后续可能继续更新本文。如果转载，请务必复制本条信息！

原文地址：http://blog.csdn.net/ebowtang/article/details/51581228

原作者博客：http://blog.csdn.net/ebowtang

本博客LeetCode题解索引：/article/3664871.html