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: "23-45"
(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]
一刷
题解：recursion
运算符两边的子字符串则为该问题的子问题。然后把子问题的解一一做运算。
public class Solution {
public List<Integer> diffWaysToCompute(String input) {
List<Integer> res = new LinkedList<>();
for(int i=0; i<input.length(); i++){
char ch = input.charAt(i);
if(ch == '-' || ch == '*' || ch == '+'){
String part1 = input.substring(0, i);
String part2 = input.substring(i+1);
List<Integer> part1Res = diffWaysToCompute(part1);
List<Integer> part2Res = diffWaysToCompute(part2);
for(int p1 : part1Res){
for(int p2 : part2Res){
int c = 0;
switch(ch){
case '+': c = p1 + p2;
break;
case '-': c = p1 - p2;
break;
case '*': c = p1 * p2;
break;
}
res.add(c);
}
}
}
}
if(res.size() == 0){
res.add(Integer.valueOf(input));
}
return res;
}
}
