LeetCode Different Ways to Add Parentheses
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题目描述
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]
题目解答
解题思路
分治法
对于输入字符串,若其中有运算符,则将其分为两部分,分别递归计算其值,然后将左值集合与右值集合进行交叉运算,将运算结果放入结果集中;若没有运算符,则直接将字符串转化为整型数放入结果集中。
代码实现
public class Solution { //缓存 Map<String, List<Integer>> map = new HashMap<>(); public List<Integer> diffWaysToCompute(String input) { if(input == null) return new ArrayList<>(); if(map.containsKey(input)) return map.get(input); int lenOfInput = input.length(); List<Integer> result = new ArrayList<>(); for(int i = 0; i < lenOfInput; i++) { char c = input.charAt(i); if(c == '+' || c == '-' || c == '*') { List<Integer> lefts = diffWaysToCompute(input.substring(0, i)); List<Integer> rights = diffWaysToCompute(input.substring(i+1)); for(int left : lefts) { for(int right : rights) { switch(c) { case '+': result.add(left+right); break; case '-': result.add(left-right); break; case '*': result.add(left*right); break; default: } } } } } if(result.isEmpty()) result.add(Integer.parseInt(input)); map.put(input, result); return result; }}
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