zoukankan      html  css  js  c++  java
  • 【LeetCode】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]

    题解:

    Solution 1 ()

    class Solution {
    public:
        vector<int> diffWaysToCompute(string input) {
            int n = input.size();
            vector<int> res;
            for(int i=0; i<n; ++i) {
                char cur = input[i];
                if(cur == '+' || cur == '-' || cur == '*') {
                    vector<int> res1 = diffWaysToCompute(input.substr(0,i));
                    vector<int> res2 = diffWaysToCompute(input.substr(i+1));
                    for(auto n1 : res1) {
                        for(auto n2 : res2) {
                            if(cur == '+') res.push_back(n1 + n2);
                            else if(cur == '-') res.push_back(n1-n2);
                            else  res.push_back(n1*n2);
                        }
                    }
                }
            }
            if(res.size() == 0)
                res.push_back(atoi(input.c_str()));
            return res;
        }
    };

    Solution 2 ()

    class Solution {
    public:
        vector<int> diffWaysToCompute(string input) {
            unordered_map<string, vector<int>> dpMap;
            return computeWithDP(input, dpMap);
        }
    
        vector<int> computeWithDP(string input, unordered_map<string, vector<int>> &dpMap) {
            vector<int> result;
            int size = input.size();
            for (int i = 0; i < size; i++) {
                char cur = input[i];
                if (cur == '+' || cur == '-' || cur == '*') {
                    // Split input string into two parts and solve them recursively
                    vector<int> result1, result2;
                    string substr = input.substr(0, i);
                    // check if dpMap has the result for substr
                    if (dpMap.find(substr) != dpMap.end())
                        result1 = dpMap[substr];
                    else
                        result1 = computeWithDP(substr, dpMap);
    
                    substr = input.substr(i + 1);
                    if (dpMap.find(substr) != dpMap.end())
                        result2 = dpMap[substr];
                    else
                        result2 = computeWithDP(substr, dpMap);
                    
                    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);
                        }
                    }
                }
            }
            // if the input string contains only number
            if (result.empty())
                result.push_back(atoi(input.c_str()));
            // save to dpMap
            dpMap[input] = result;
            return result;
        }
    };
  • 相关阅读:
    表单全选,不选和反选
    利用js改变宽,高等属性
    点击显示与隐藏
    改变div的不同属性
    给三个不同的div变色
    经典排序算法实现
    基本排序算法的实现
    排序的概念及分类实现
    #和##运算符实例
    #pragma预处理实例
  • 原文地址:https://www.cnblogs.com/Atanisi/p/6810192.html
Copyright © 2011-2022 走看看