241. Different Ways to Add Parentheses
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LeetCode
- 题目地址:https://leetcode.com/problems/different-ways-to-add-parentheses/#/description
问题描述&解题思路:给一个字符串,比如”2*3-4*5“,根据运算顺序的不同组合,可以得到以下结果:
(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
一开始没什么思路,看到答案用递归的方法,以符号作为分割,左边字符串返回一组int向量,右边也是,再两两匹配,做+,-,*(根据该分割的符号来做)。递归的终止条件是字符串没有运算符,这时候返回一个数。
但是递归的方法,会计算很多重复的部分(但是这道题的测试集不大,用动态规划的记忆的方法,反而慢了一点),于是参考动态规划,用一个hash map来存储已经做过递归的结果,其中因为有两个数字,一个是字符串起始位置,一个是结束位置,于是做了一个转换begin*size + end,这样来做索引。
递归代码
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 == '*') { // Split input string into two parts and solve them recursively vector<int> result1 = diffWaysToCompute(input.substr(0, i)); vector<int> result2 = diffWaysToCompute(input.substr(i+1)); 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())); return result; }};
- dp代码:
class Solution {public: vector<int> diffWaysToCompute(string input) { unordered_map<int,vector<int>> m; return dp(input,0,input.size(),m); } //e: from 0 to size - 1 int trans(int b, int e, const int& size) { return b * size + e; } //from b to e-1 vector<int> dp(const string& input, int b, int e, unordered_map<int,vector<int>> m) { vector<int> res; int size = input.size(); for (int i = b; i < e; i++) { //每有一个运算符,就做一次 if (input[i] == '+' || input[i] == '-' || input[i] == '*') { int l = trans(b,i-1,size); int r = trans(i+1,e-1,size); vector<int> r1,r2; //left if (m.count(l) == 0) { m[l] = dp(input,b,i,m); } r1 = m[l]; //right if (m.count(r) == 0) { m[r] = dp(input,i+1,e,m); } r2 = m[r]; for (auto& a : r1) { for (auto& b : r2) { if (input[i] == '+') { res.push_back(a+b); } else if (input[i] == '-') { res.push_back(a-b); } else if (input[i] == '*') { res.push_back(a*b); } } } } } //如果res是空的,那么只有数字没有运算符 if (res.empty()) { res.push_back(atoi(input.substr(b,e-b).c_str())); } return res; }};
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