难度: Medium
原题连接
内容描述
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"
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
思路 1 - 时间复杂度: O(N)- 空间复杂度: O(N)******
递归
beats 84.41%
class Solution:
def diffWaysToCompute(self, input):
"""
:type input: str
:rtype: List[int]
"""
if input.isdigit():
return [int(input)]
res = []
for i in range(len(input)):
if input[i] in "-+*":
res1 = self.diffWaysToCompute(input[:i])
res2 = self.diffWaysToCompute(input[i+1:])
res.extend(self.helper(j, k, input[i]) for j in res1 for k in res2)
return res
def helper(self, m, n, op):
if op == "+":
return m + n
elif op == "-":
return m - n
else:
return m * n