LeetCode——Submission Details

Description：

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]

Credits:
Special thanks to @mithmatt for adding this problem and creating all test cases.

使用分治递归的方法首先将问题划分成子问题，再对子问题的解进行全排列合并，最终得到问题的解。

深刻理解递归还是很有必要的。

public class Solution {
    public List<Integer> diffWaysToCompute(String input) {

        List<Integer> resList = new ArrayList<Integer>();

        for(int i=0; i<input.length(); i++) {
            char op = input.charAt(i);
            if(op == '+' || op == '-' || op == '*') {

                List<Integer> leftList = diffWaysToCompute(input.substring(0, i));
                List<Integer> rightList = diffWaysToCompute(input.substring(i+1));

                for(int left : leftList) {
                    for(int right : rightList) {
                        if(op == '+') {
                            resList.add(left + right);
                         }
                        else if(op == '-') {
                            resList.add(left - right);
                        }
                        else {
                            resList.add(left * right);
                        }
                    }
                }

            }
        }

        if(resList.size() == 0) {
            resList.add(Integer.parseInt(input));
        }

        return resList;
    }
}

这题竟然没有不是个位数的case。