Different Ways to Add Parentheses

问题描述

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]

 

解决思路

其实挺巧妙的，主要参考https://leetcode.com/discuss/48488/c-4ms-recursive-method

理解这句话很重要：The key idea for this solution is: each operator in this string could be the last operator to be operated.

核心思想是递归，然后边界条件是输入的字符串仅为数字。

 

程序

public class DifferentWaysToCompute {
	public List<Integer> diffWaysToCompute(String input) {
		List<Integer> list = new ArrayList<Integer>();
		for (int i = 0; i < input.length(); i++) {
			char c = input.charAt(i);
			if (c == '+' || c == '-' || c == '*') {
				// Split input string into two parts and solve them recursively
				List<Integer> left = diffWaysToCompute(input.substring(0, i));
				List<Integer> right = diffWaysToCompute(input.substring(i+1));
				
				// calculate each part
				for (int j = 0; j < left.size(); j++) {
					for (int k = 0; k < right.size(); k++) {
						if (c == '+') {
							list.add(left.get(j) + right.get(k));
						} else if (c == '-') {
							list.add(left.get(j) - right.get(k));
						}else {
							list.add(left.get(j) * right.get(k));
						}
					}
				}
			}
		}
		
		if (list.isEmpty()) {
			// input string contains only number
			list.add(Integer.valueOf(input));
		}
		
		return list;
	}
}