Leetcode 241. Different Ways to Add Parentheses
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题目描述:
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]
题目分析:
这是一道分治算法的题目,要算出所有不同优先顺序下算式的结果,那么根据分治算法的思想,将问题转化为子问题求解,则应该在遇到一个运算符后将前后两部分算式分别递归求结果,然后再合并,得到最终的算式结果。
具体步骤:
- step1:遍历整个字符串,遇到运算符后,以运算符为分界,对前后两个子串分别递归求运算结果。
- step2:分治结束后,对得到的两个子串的各个不同的运算结果一一对应进行合并。
- step3:如果没有运算符,说明只有一个数字,则将其放入数组中返回即可。
代码:
#include <iostream>#include <string>#include <sstream>#include <vector>using namespace std;class Solution {public: vector<int> diffWaysToCompute(string input) { vector<int> res; for (int i = 0; i < input.size(); i++){ if (input[i] == '+' || input[i] == '-' || input[i] == '*'){ /*对前后两个子串分别递归求运算结果*/ vector<int> res1 = diffWaysToCompute(input.substr(0, i)); vector<int> res2 = diffWaysToCompute(input.substr(i + 1)); /*对两个子串各个不同的运算结果一一对应进行合并*/ for (int j = 0; j < res1.size(); j++){ for (int k = 0; k < res2.size(); k++){ if (input[i] == '+'){ res.push_back(res1[j] + res2[k]); } else if (input[i] == '-'){ res.push_back(res1[j] - res2[k]); } else{ res.push_back(res1[j] * res2[k]); } } } } } /*如果没有运算符,说明只有一个数字,则将其放入数组中返回即可*/ if (res.empty()){ int temp; stringstream ss; ss << input; ss >> temp; res.push_back(temp); } return res; }};
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