[LeetCode] Different Ways to Add Parentheses

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[LeetCode] Different Ways to Add Parentheses

楚兴 2015-11-09 18:08:00 浏览891
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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]

解题思路

分而治之。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。

实现代码

Java:

// Runtime: 6 ms
public class Solution {
    public List<Integer> diffWaysToCompute(String input) {
        List<Integer> res = new ArrayList<Integer>();
        for (int i = 0; i < input.length(); i++) {
            char ch = input.charAt(i);
            if (ch == '+' || ch == '-' || ch == '*') {
                List<Integer> left = diffWaysToCompute(input.substring(0, i));
                List<Integer> right = diffWaysToCompute(input.substring(i + 1));
                for (int n : left) {
                    for (int m : right) {
                        switch (ch) {
                        case '+':
                            res.add(n + m);
                            break;
                        case '-':
                            res.add(n - m);
                            break;

                        case '*':
                            res.add(n * m);
                            break;
                        }
                    }
                }
            }
        }

        if (res.size() == 0) {
            res.add(Integer.parseInt(input));
        }

        return res;
    }
}

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