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]

思路:就是一个递归求和的问题，有点归并的感觉

代码如下(已通过leetcode)

public class Solution {

public List<Integer> diffWaysToCompute(String input) {

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

if(input==null ||input.length()==0) return list;

if(!input.contains("+")&&!input.contains("-")&&!input.contains("*")) {

list.add(Integer.valueOf(input));

return list;

}

for(int i=0;i<input.length();i++) {

char ops=input.charAt(i);

if(ops=='+'||ops=='-'||ops=='*') {

List<Integer> left=diffWaysToCompute(input.substring(0,i));

List<Integer> right=diffWaysToCompute(input.substring(i+1,input.length()));

for(int leftvalue:left) {

for(int rightvalue:right) {

switch(ops) {

case '+': list.add(leftvalue+rightvalue);

break;

case '-': list.add(leftvalue-rightvalue);

break;

case '*': list.add(leftvalue*rightvalue);

break;

}

}

}

}

}

return list;

}

}