[Leetcode] 241. Different Ways to Add Parentheses 解题报告
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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]
思路:
我开始的解法是回溯,发现代码实在是太复杂了,思路也不清晰。后来发现这道题目的正解是分治。不得不说,下面的分治策略的实现还是非常巧妙的:
题目的实质就是看运算符有多少种排列,然后每种排列都会对应一个最终结果。我们的做法是:对于每一个运算符,我们首先计算出它的左侧和右侧分别有多少种运算结果,然后把它们组合一下,再用该运算符进行运算获得结果。这里面会用到递归的实现方法,所以需要注意递归的边界(就是整个字符串就是一个数组,没有运算符)。
代码:
class Solution {public: vector<int> diffWaysToCompute(string input) { return diffWaysToCompute(input, 0, input.length()); }private: vector<int> diffWaysToCompute(string& input, int start, int end) { vector<int> ret; char c; for(int i = start; i < end; ++i) { c = input[i]; if(c == '+' || c == '-' || c == '*') { vector<int> left = diffWaysToCompute(input, start, i); vector<int> right = diffWaysToCompute(input, i + 1, end); for(auto it1 = left.begin(); it1 != left.end(); ++it1) { for(auto it2 = right.begin(); it2 != right.end(); ++it2) { if(c == '+') { ret.push_back(*it1 + *it2); } else if(c == '-') { ret.push_back(*it1 - *it2); } else if(c == '*') { ret.push_back(*it1 * *it2); } } } } } if(ret.size() == 0) { string substr = input.substr(start, end - start); ret.push_back(atoi(substr.c_str())); } return ret; }};
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