题目:
时间复杂度O(n2)
class Solution:
def twoSum(self, nums: List[int], target: int) -> List[int]:
for i in range(len(nums)):
for j in range(i+1,len(nums)):
if nums[i] + nums[j] == target:
return [i, j]
但注意,不要用enumerate函数写,会超时:
class Solution:
def twoSum(self, nums: List[int], target: int) -> List[int]:
size = len(nums)
for i, m in enumerate(nums):
j = i+1
while j < size :
if nums[i] + nums[j] == target:
return [i, j]
else:
j+=1
python大法好:用list+in方法,只需要一个for循环就能解决问题了(但其实是python的in帮我们做了一个查找的循环)
时间复杂度O(n2)
class Solution:
def twoSum(self, nums: List[int], target: int) -> List[int]:
for i in range(len(nums)):
if target-nums[i] in nums:
if i != nums.index(target-nums[i]):
return [i, nums.index(target-nums[i])]
时间复杂度O(n)
class Solution:
def twoSum(self, nums: List[int], target: int) -> List[int]:
_dict = {}
for i, m in enumerate(nums):
_dict[m] = i
for i, m in enumerate(nums):
j = _dict.get(target - m)
if j is not None and i != j:
return [i, j]
时间复杂度O(n)
class Solution:
def twoSum(self, nums: List[int], target: int) -> List[int]:
d = {}
for i in range(len(nums)):
a = target - nums[i]
if nums[i] in d:
return d[nums[i]],i
else:
d[a] = i
自己想想写写就明白了,字典d里键值对 {k:v}的含义是,与k能凑成target的值在nums中的位置为v。(即nums[i]=k时,nums[v]+num[i]=target。)
边在字典中记下互补这个位置(value)所需互补数(key)边遍历nums数组,之后的遇到nums[i]=之前记录的某个互补数时就是找到了,返回他的位置(value)和 i 就完成了。
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