0241. Different Ways to Add Parentheses
https://leetcode.com/problems/different-ways-to-add-parentheses
Description
Given a string expression
of numbers and operators, return all possible results from computing all the different possible ways to group numbers and operators. You may return the answer in any order.
Example 1:
**Input:** expression = "2-1-1"
**Output:** [0,2]
**Explanation:**
((2-1)-1) = 0
(2-(1-1)) = 2
Example 2:
**Input:** expression = "2*3-4*5"
**Output:** [-34,-14,-10,-10,10]
**Explanation:**
(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
Constraints:
1 <= expression.length <= 20
expression
consists of digits and the operator'+'
,'-'
, and'*'
.All the integer values in the input expression are in the range
[0, 99]
.
ac: divide and conquer
class Solution {
public List<Integer> diffWaysToCompute(String input) {
List<Integer> res = new ArrayList<Integer>();
List<Integer> l1 = new ArrayList<Integer>();
List<Integer> l2 = new ArrayList<Integer>();
for (int i = 0; i < input.length(); i++) {
// meet operator, divide 2 parts, get result list
if (input.charAt(i) == '+'
|| input.charAt(i) == '-'
|| input.charAt(i) == '*') {
l1 = diffWaysToCompute(input.substring(0, i));
l2 = diffWaysToCompute(input.substring(i+1));
// conquer 2 result
for (int i1 : l1) {
for (int i2 : l2) {
int tmp = 0;
switch(input.charAt(i)) {
case '+':
tmp = i1 + i2;
break;
case '-':
tmp = i1 - i2;
break;
case '*':
tmp = i1 * i2;
}
res.add(tmp);
}
}
}
}
// do not contain operator, return value
if (res.size() == 0) res.add(Integer.parseInt(input));
return res;
}
}
/**
walk, meet + - *, divide 2 parts, recursion
**/
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