- 分类：分治
- 时间复杂度: O(nlogn)
241. 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"
Output: [0, 2]
Explanation:
((2-1)-1) = 0
(2-(1-1)) = 2
Example 2:
Input: "2*3-4*5"
Output: [-34, -14, -10, -10, 10]
Explanation:
(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
代码：
分治方法:
class Solution:
def diffWaysToCompute(self, input):
"""
:type input: str
:rtype: List[int]
"""
ans=[]
if input==None and len(input)==0:
return ans
return self.getResult(input)
def getResult(self, input):
if ("+" not in input) and ("-" not in input) and ("*" not in input):
return [int(input)]
res=[]
for i in range(len(input)):
if not input[i].isdigit():
res+=self.decare(self.getResult(input[:i]),self.getResult(input[i+1:]),input[i])
return res
def decare(self,array1,array2,op):
res=[]
for a in array1:
for b in array2:
if op=="+":
res.append(a+b)
elif op=="-":
res.append(a-b)
else:
res.append(a*b)
return res
讨论：
1.一开始看到这道题的时候，看见括号，还以为要用stack。听完讲解之后发现还是用分治，说不定还能用树做？？？
2.知识点：分治+笛卡尔集
241
