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.
The test cases are generated such that the output values fit in a 32-bit integer and the number of different results does not exceed 104.
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 <= 20expression consists of digits and the operator '+', '-', and '*'.[0, 99].'-' or '+' denoting the sign.class Solution {
public List<Integer> diffWaysToCompute(String expression) {
int n = expression.length();
List<Integer>[][] memo = new List[n][n];
return solve(expression, memo, 0, n - 1);
}
List<Integer> solve(String expr, List<Integer>[][] memo, int lt, int rt) {
if (memo[lt][rt] != null)
return memo[lt][rt];
if (lt == rt)
return memo[lt][rt] = new ArrayList<>(List.of(expr.charAt(lt) - '0'));
else if (rt - lt == 1)
return memo[lt][rt] = new ArrayList<>(List.of(Integer.parseInt(expr.substring(lt, rt + 1))));
List<Integer> res = new ArrayList<>();
for (int i = lt; i <= rt; i++) {
char c = expr.charAt(i);
if (Character.isDigit(c))
continue;
List<Integer> ltOperand = solve(expr, memo, lt, i - 1);
List<Integer> rtOperand = solve(expr, memo, i + 1, rt);
// System.out.println(expr.substring(0, i) + "\t" + expr.substring(i + 1));
// System.out.println("ltOperands\t" + ltOperand);
// System.out.println("rtOperands\t" + rtOperand);
for (int l : ltOperand)
for (int r : rtOperand) {
// System.out.println(l + "\t" + r);
res.add(switch (c) {
case '+' -> l + r;
case '-' -> l - r;
case '*' -> l * r;
default -> 0;
});
}
}
return memo[lt][rt] = res;
}
}