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_241.java
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package com.fishercoder.solutions;
import java.util.LinkedList;
import java.util.List;
/**
* 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]
*/
public class _241 {
public static class Solution1 {
/**Time: O(n * 4^n / n^(3/2)) ~= n * Catalan numbers = n * (C(2n, n) - C(2n, n - 1)),
due to the size of the results is Catalan numbers,
and every way of evaluation is the length of the string,
so the time complexity is at most n * Catalan numbers.
Space: O(n * 4^n / n^(3/2)), the cache size of lookup is at most n * Catalan numbers.*/
/**
* Credit: https://discuss.leetcode.com/topic/19901/a-recursive-java-solution-284-ms
*/
public List<Integer> diffWaysToCompute(String input) {
List<Integer> answer = new LinkedList<>();
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> answer1 = diffWaysToCompute(part1);
List<Integer> answer2 = diffWaysToCompute(part2);
for (int a1 : answer1) {
for (int a2 : answer2) {
int result = 0;
switch (input.charAt(i)) {
case '+':
result = a1 + a2;
break;
case '-':
result = a1 - a2;
break;
case '*':
result = a1 * a2;
break;
default:
break;
}
answer.add(result);
}
}
}
}
if (answer.size() == 0) {
answer.add(Integer.valueOf(input));
}
return answer;
}
}
}