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 = 0; i < length; ++i){
            char c = input[i];
            if(c == '+' || c == '-' || c == '*'){
                string inputLeft = input.substr(0, i);
                string inputRight = input.substr(i+1);
                vector<int>leftResult = diffWaysToCompute(inputLeft);
                vector<int>rightResult = diffWaysToCompute(inputRight);
                for(int j = 0; j < leftResult.size(); ++j){
                    for(int k = 0; 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;
    }
}