LeetCode - Different Ways to Add Parentheses
Problem description
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:
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Input: "2-1-1"
Output: [0, 2]
Explanation:
((2-1)-1) = 0
(2-(1-1)) = 2
Example 2:
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Input: "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
Analysis
The idea is to use divide and conquer.
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class Solution {
public List<Integer> diffWaysToCompute(String input) {
List<Integer> ret = new LinkedList<Integer>();
for (int i=0; i<input.length(); 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;
break;
}
ret.add(c);
}
}
}
}
if (ret.size() == 0) {
ret.add(Integer.valueOf(input));
}
return ret;
}
}
What to improve
- learn how to analysis this type of question.