与C++想法一样,个人实现的是最简单的for循环:
class Solution:
def twoSum(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: List[int]
"""
towSum=[]
for i in range(len(nums)):
for j in range(i+1,len(nums)):
if(nums[i]+nums[j]==target):
towSum.append(i)
towSum.append(j)
return towSum
以下优化转自(wx公众号:小詹学python[先注明:不是博主的公众号]):
1.将上述双循环可以优化一下为单循环:
class Solution:
def twoSum(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: List[int]
"""
towSum=[]
for i in range(len(nums)):
onenum=nums[i]
twonum=target-nums[i]
if twonum in nums:
j=nums.index(twonum)
if i!=j:
towSum.append(i)
towSum.append(j)
return towSum
上述两种方法循环时间都很长:
按照官方代码来说,还是采用字典更好:
2.字典优化:
class Solution:
def twoSum(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: List[int]
"""
#创建字典一:存入nums[i],i
num_dict={num[i]: i fori in range(len(nums))}
#创建字典二:存入i:target-nums[i]
num_dict2={i:target-num[i] fori in range(len(nums))}
towSum=[]
for i in range(len(nums)):
j=num_dict.get(num_dict2.get(i))
if (j is not None) and (j!=i):
towSum=[i,j]
break
return towSum