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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]
public class Solution { public List<Integer> diffWaysToCompute(String input) { //递归思想，遇到运算符，就对左右两边进行递归，然后对结果进行组合 List<Integer> res=new ArrayList<Integer>(); for(int i=0;i<input.length();i++){ char c=input.charAt(i); if(!Character.isDigit(c)){ List<Integer> left=diffWaysToCompute(input.substring(0,i)); List<Integer> right=diffWaysToCompute(input.substring(i+1)); for(int k=0;k<left.size();k++){ for(int j=0;j<right.size();j++){ switch(c){ case ‘+‘: res.add(left.get(k)+right.get(j)); break; case ‘-‘: res.add(left.get(k)-right.get(j)); break; case ‘*‘: res.add(left.get(k)*right.get(j)); break; } } } } } if(res.isEmpty()) res.add(Integer.valueOf(input));//注意单个数字的输入，比如：1，此时需要特殊判断 return res; } }
[leedcode 241] Different Ways to Add Parentheses
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原文地址：http://www.cnblogs.com/qiaomu/p/4714253.html
