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.
解法思想：递归
代码如下：
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;
}
}
运行结果：
