LeetCode OJ ： Different Ways to Add Parentheses（在不同位置增加括号的方法）

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

 class Solution {
 public:
     vector<int> diffWaysToCompute(string input) {
         vector<int> result;
         int length = input.length();
         for(int i = ; i < length; ++i){
             char c = input[i];
             if(c == '+' || c == '-' || c == '*'){
                 string inputLeft = input.substr(, i);
                 string inputRight = input.substr(i+);
                 vector<int>leftResult = diffWaysToCompute(inputLeft);
                 vector<int>rightResult = diffWaysToCompute(inputRight);
                 for(int j = ; j < leftResult.size(); ++j){
                     for(int k = ; k < rightResult.size(); ++k){
                         if(c == '+')
                             result.push_back(leftResult[j] + rightResult[k]);
                         else if(c == '-')
                             result.push_back(leftResult[j] - rightResult[k]);
                         else if(c == '*')
                             result.push_back(leftResult[j] * rightResult[k]);
                     }
                 }
             }
         }
         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
             result.push_back(stoi(input));
         return result;
     }
 };

java版本如下所示：

 public class Solution {
     public List<Integer> diffWaysToCompute(String input) {
         ArrayList<Integer> ret = new ArrayList<Integer>();
         for(int i = 0; i < input.length(); ++i){
             char c = input.charAt(i);
             if(c == '+' || c == '-' || c == '*'){
                 List<Integer> leftRet = diffWaysToCompute(input.substring(0, i));
                 List<Integer> rightRet = diffWaysToCompute(input.substring(i+1, input.length()));
                 for(int left : leftRet){
                     for(int right : rightRet){
                         if(c == '+'){
                             ret.add(left + right);
                         }else if(c == '-'){
                             ret.add(left - right);
                         }else{
                             ret.add(left * right);
                         }
                     }
                 }
             }
         }
         if(ret.isEmpty()){
             ret.add(Integer.parseInt(input));
         }
         return ret;
     }
 }