[leetcode]different ways to add parentheses

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”.

1 2	((2-1)-1) = 0 (2-(1-1)) = 2

Output: [0, 2]

Example 2

Input: “23-45”

(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]

代码

这道题跟Unique Binary Search Trees II思路类似，都是利用分治的思想，可对照理解

public class Solution {
    public List<Integer> diffWaysToCompute(String input) {
        if(input == null || input.length() == 0) return new ArrayList<Integer>();
        
        List<Integer> res = new ArrayList<>();
        
        for(int i = 0; i < input.length(); i++){
            char ch = input.charAt(i);
            //字符ch为+、-、*，可以采用分治的方法将input在ch分为左右两部分，递归计算
            if(ch == '-' || ch == '+' || ch == '*'){
                List<Integer> leftList = diffWaysToCompute(input.substring(0, i));
                List<Integer> rightList = diffWaysToCompute(input.substring(i+1));
                
                for(int left : leftList){
                    for(int right : rightList){
                        switch(ch){
                            case '-': res.add(left - right);
                                break;
                            case '+': res.add(left + right);
                                break;
                            case '*': res.add(left * right);
                                break;
                        }
                    }
                }
            }
        }
        //数字
        if(res.size() == 0) res.add(Integer.valueOf(input));
        
        return res;
    }
}

[leetcode]different ways to add parentheses

题目描述

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