241. Different Ways to Add Parentheses

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.

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DFS算法

public class Solution {    public List<Integer> diffWaysToCompute(String input) {        List<Integer> ret=new LinkedList<Integer>();        int len=input.length();        for(int i=0;i<len;i++){            if(input.charAt(i)=='-'||               input.charAt(i)=='*'||               input.charAt(i)=='+'){                   String part1=input.substring(0,i);                   String part2=input.substring(i+1);                   List<Integer> part1Ret=diffWaysToCompute(part1);                   List<Integer> part2Ret=diffWaysToCompute(part2);                   for(Integer p1:part1Ret){                       for(Integer p2:part2Ret){                           int c=0;                           switch(input.charAt(i)){                               case '+':c=p1+p2;                                   break;                               case '-':c=p1-p2;                                   break;                               case '*': c=p1*p2;                                  }                           ret.add(c);                       }                   }               }        }        if(ret.size()==0){            ret.add(Integer.valueOf(input));        }        return ret;    }}