Different Ways to Add Parentheses

標籤：

問題描述

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]

 

解決思路

其實挺巧妙的，主要參考https://leetcode.com/discuss/48488/c-4ms-recursive-method

理解這句話很重要：The key idea for this solution is: each operator in this string could be the last operator to be operated.

核心思想是遞迴，然後邊界條件是輸入的字串僅為數字。

 

程式

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 recursivelyList<Integer> left = diffWaysToCompute(input.substring(0, i));List<Integer> right = diffWaysToCompute(input.substring(i+1));// calculate each partfor (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 numberlist.add(Integer.valueOf(input));}return list;}}

　　

Different Ways to Add Parentheses