leetcode题解c++ | 241. Different Ways to Add Parentheses

题目：

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]




分析：

这是一道分治的题目。其核心思想为，对于每个式子，可根据一个操作运算符（以+为例），分为左右两段，假设左边的可能值有x种，右边的可能值有y种，则左右相加能得到x*y种可能性。对左右各递归地操作。遍历式中的所有操作运算符，则可以得到所有的可能情况。

c++题解：

class Solution {public:vector<int> compute(string s, int l, int r){    vector<int> left,right,ans;    bool tag = 0;    for(int i=l; i<=r; ++i)    {        if(s[i]>='0' && s[i]<='9')            continue;        else        {            tag = 1;            left = compute(s,l,i-1);            right = compute(s,i+1,r);            vector<int>::iterator it_left = left.begin();            while(it_left != left.end())            {                vector<int>::iterator it_right = right.begin();                while(it_right != right.end())                {                    if(s[i]=='+')                        ans.push_back(*it_left+(*it_right));                    else if(s[i]=='-')                        ans.push_back(*it_left-(*it_right));                    else                        ans.push_back(*it_left*(*it_right));                    ++it_right;                }                ++it_left;            }        }    }    if(tag==0)    {        int n=0;        for(int i=l; i<=r; ++i)            n = n*10 + (s[i]-'0');        ans.push_back(n);    }    return ans;}vector<int> diffWaysToCompute(string input){    return compute(input, 0, input.length()-1);}};