Leetcode: Different Ways to Add Parentheses
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]
用到了Divide and Conquer, 跟 Leetcode: Unique Binary Search Trees II 很像
在input string里遍历各个operator, 依据每个operator分成左右子串,左右子串做递归返回所有可能的results,然后全排列。
注意很巧妙的一点在于寻找operator来划分,然后如果像20-23行那样没有划分成功(因为每找到operator)则Inter.parseInt(input),这样省去了计算多位integer的工夫。比如input.charAt(i)=2, input.charAt(i+1)=3, 我们不需要再自己计算得到23.
1 public class Solution { 2 public List<Integer> diffWaysToCompute(String input) { 3 List<Integer> res = new ArrayList<Integer>(); 4 if (input==null || input.length()==0) return res; 5 6 for (int i=0; i<input.length(); i++) { 7 char c = input.charAt(i); 8 if (!isOperator(c)) continue; 9 List<Integer> left = diffWaysToCompute(input.substring(0, i)); 10 List<Integer> right = diffWaysToCompute(input.substring(i+1)); 11 for (int j=0; j<left.size(); j++) { 12 for (int k=0; k<right.size(); k++) { 13 int a = left.get(j); 14 int b = right.get(k); 15 res.add(calc(a,b,c)); 16 } 17 } 18 } 19 20 if (res.size() == 0) { //input is not null or size==0, but res.size()==0, meaning input is just a number 21 res.add(Integer.parseInt(input)); 22 } 23 return res; 24 } 25 26 public boolean isOperator(char c) { 27 if (c=='+' || c=='-' || c=='*') return true; 28 return false; 29 } 30 31 public int calc(int a, int b, char c) { 32 int res = Integer.MAX_VALUE; 33 switch(c) { 34 case '+': res = a+b; break; 35 case '-': res = a-b; break; 36 case '*': res = a*b; break; 37 } 38 return res; 39 } 40 }