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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]
这里的主要思想是讲左右的子串分别算出来之后，再做全排列就行了，可能有很所重复计算所以说效率比较低
1 class Solution { 2 public: 3 vector<int> diffWaysToCompute(string input) { 4 vector<int> result; 5 int length = input.length(); 6 for(int i = 0; i < length; ++i){ 7 char c = input[i]; 8 if(c == ‘+‘ || c == ‘-‘ || c == ‘*‘){ 9 string inputLeft = input.substr(0, i); 10 string inputRight = input.substr(i+1); 11 vector<int>leftResult = diffWaysToCompute(inputLeft); 12 vector<int>rightResult = diffWaysToCompute(inputRight); 13 for(int j = 0; j < leftResult.size(); ++j){ 14 for(int k = 0; k < rightResult.size(); ++k){ 15 if(c == ‘+‘) 16 result.push_back(leftResult[j] + rightResult[k]); 17 else if(c == ‘-‘) 18 result.push_back(leftResult[j] - rightResult[k]); 19 else if(c == ‘*‘) 20 result.push_back(leftResult[j] * rightResult[k]); 21 } 22 } 23 } 24 } 25 if(result.empty())//这一步主要的作用是讲分治得到的最后的字符转换成数字 26 result.push_back(stoi(input)); 27 return result; 28 } 29 };
LeetCode OJ ： Different Ways to Add Parentheses（在不同位置增加括号的方法）
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原文地址：http://www.cnblogs.com/-wang-cheng/p/4853883.html
