分治和自顶向下动态规划算法

来源：互联网发布：人工智能框架编辑：程序博客网时间：2024/09/21 09:17

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]

解题的思路是分治算法，比如上面的2*3-4*5

让str1 = 2 *3，str2 = 4 × 5；如此，变为str1 - str2，把一个问题分解成了两个规模更小的子问题。

自顶向下动态规划算法：用一个hash表保存问题与解集的映射，如果次问题已经出现过，直接查找hash表，反之求解，并把解集放如hash表。

#include<vector>#include<string>#include<unordered_map>using namespace std;class Solution {    unordered_map<string,vector<int>> map;public:    vector<int> diffWaysToCompute(string input) {        return divide(input);    }    vector<int> divide(const string& str){        if(map.find(str) != map.end())            return map[str];        vector<int> res;        for(int i = 0; i < str.size(); ++i){            char ch = str[i];            if(isOpperator(ch)){                string sub1 = str.substr(0,i);                string sub2 = str.substr(i+1);                vector<int> v1 = divide(sub1);                vector<int> v2 = divide(sub2);                vector<int> v = genrate(v1,v2,ch);                for(int each: v)                    res.push_back(each);            }        }        if(res.empty())            res.push_back(atoi(str.data()));        map[str] = res;        return res;    }    bool isOpperator(char c){        return c == '+' || c == '-' || c =='*';    }    vector<int> genrate(const vector<int>& a,const vector<int>& b,char c){        vector<int> res;        for(int i = 0; i < a.size(); ++i){            for(int j = 0; j <b.size(); ++j){                int val;                if(c == '+')                    val = a[i] + b[j];                else if(c =='-')                    val = a[i] - b[j];                else if(c == '*')                    val = a[i] * b[j];                res.push_back(val);            }        }        return res;    }};

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