241. Different Ways to Add Parentheses
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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]
题解:
使用divide and conquer的思想,对input进行遍历,每找到一个operator,就把input分成两份,part1 (operator) part2;对part1和part2分别计算,最后再把两个部分得出的值合起来运算。这个过程会存在一些重复运算,可以使用dp memoization的思想,把计算过的值存储下来,下次可以重复利用。
代码:
public class Solution { public List<Integer> diffWaysToCompute(String input) { Map<String, List<Integer>> map = new HashMap<>(); List<Integer> result = computeWithMemo(input, map); return result; } private List<Integer> computeWithMemo(String input, Map<String, List<Integer>> map) { List<Integer> result = new LinkedList<>(); 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> result1; List<Integer> result2; if(map.containsKey(part1)) { result1 = map.get(part1); } else { result1 = computeWithMemo(part1, map); } if(map.containsKey(part2)) { result2 = map.get(part2); } else { result2 = computeWithMemo(part2, map); } for(Integer o1 : result1) { for(Integer o2: result2) { result.add(helper(o1,o2,input.charAt(i))); } } } } if(result.size() == 0) { result.add(Integer.valueOf(input)); } map.put(input, result); return result; } private int helper(int o1, int o2, char c) { if(c == '+') { return o1 + o2; } else if(c == '*') { return o1 * o2; } else { return o1 - o2; } }}
时间复杂度:
假如不使用map记录计算过的值:
f(n) = f(1)+f(2)+..+f(n-1)+f(n-1)+f(n-2)..+f(1) = 2*(f(1)+f(2)+...+f(n-1))=> f(n+1) = 2*(f(1)+f(2)+..+f(n)) = 2*(f(1)+f(2)+...+f(n-1)) + 2f(n) = f(n)+2f(n) = 3f(n) => f(n+1) = 3f(n) = 3*3*f(n-1) = ... = 3^(n+1)Therefore, O(3^n)
题解使用map避免了重复计算,所以应该为f(n) = f(1)+f(2)+..+f(n-1) => f(n+1) = f(1)+f(2)+..+f(n) = 2*(f(1)+f(2)+...+f(n-1)) = 2f(n)=> f(n+1) = 2f(n) = 2*2*f(n-1) = ... = 2^(n+1)Therefore, O(2^n)
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