LeetCode----241. Different Ways to Add Parenthese(M)分治

1 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]

2 分析和实现

用分治思想实现

class Solution {
public:vector<int> diffWaysToCompute(string input) {vector<int> res;for(int i = 0; i < input.size(); i++) {char cur = input[i];if(cur == '+' || cur == '-' || cur == '*') {vector<int> res_left = diffWaysToCompute(input.substr(0, i));vector<int> res_right = diffWaysToCompute(input.substr(i + 1));for(int x = 0; x < res_left.size(); x++) {for(int y = 0; y < res_right.size(); y++) {if(cur == '+')res.push_back(res_left[x] + res_right[y]);else if(cur == '-')res.push_back(res_left[x] - res_right[y]);else if(cur == '*')res.push_back(res_left[x] * res_right[y]);}}}}//单独一个数字时if(res.empty()) {res.push_back(atoi(input.c_str()));}return res;   }
};

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