Different Ways to Add Parentheses

原文地址：https://leetcode.com/problems/different-ways-to-add-parentheses/description/

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]

这种问题可以使用分治法，递归去处理运算符两边的字符串，边界条件就是不再可以按照运算符分割字符串，即字符串中只包含数字。

public class DifferentWaysToAddParentheses {    public static List<Integer> diffWaysToCompute(String input) {        if (input == null)            return null;        List<Integer> res = new ArrayList<>();        for (int i = 0; i < input.length(); i++) {            if (input.charAt(i) == '+' || input.charAt(i) == '-' || input.charAt(i) == '*') {                List<Integer> res1 = diffWaysToCompute(input.substring(0, i));                List<Integer> res2 = diffWaysToCompute(input.substring(i + 1));                for (int index1 = 0; index1 < res1.size(); index1++) {                    for (int index2 = 0; index2 < res2.size(); index2++) {                        if (input.charAt(i) == '+') {                            res.add(res1.get(index1) + res2.get(index2));                        } else if (input.charAt(i) == '-') {                            res.add(res1.get(index1) - res2.get(index2));                        } else {                            res.add(res1.get(index1) * res2.get(index2));                        }                    }                }            }        }        if (res == null || res.size() == 0)            res.add(Integer.valueOf(input));        return res;    }    public static void main(String[] args) {        String input = "2*3-4*5";        System.out.println(diffWaysToCompute(input));    }}