1. Two Sum
Given an array of integers, return indices of the two numbers such that they add up to a specific target.
You may assume that each input would have exactly one solution, and you may not use the same element twice.
Example:
Given nums = [2, 7, 11, 15], target = 9,
Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].
AC code
class Solution:
def twoSum(self, nums: List[int], target: int) -> List[int]:
if nums == []:
return []
if len(nums) == 1 and nums[0] != target:
return []
elif len(nums) == 1 and nums[0] == target:
return [0]
for i in range(len(nums)):
number = target - nums[i]
if number in nums[i + 1:]:
return [i, i + nums[i + 1:].index(number) + 1]
return []
如果使用指针来遍历数组中的元素进行查找会超出时间。
class Solution:
def twoSum(self, nums: List[int], target: int) -> List[int]:
if nums == []:
return []
if len(nums) == 1 and nums[0] != target:
return []
elif len(nums) == 1 and nums[0] == target:
return [0]
p = 0
cur = 1
while cur < len(nums):
if nums[p] + nums[cur] == target:
return [p, cur]
elif nums[p] + nums[cur] != target and cur == len(nums) - 1:
p += 1
cur = p + 1
else:
cur += 1
return []
一种更省内存和运行时间的方法是利用python内置的数据结构map构建hashmap来存放数组中的元素，进而再进行后续的查找。
class Solution:
def twoSum(self,nums,target):
d = {}
n = len(nums)
for x in range(n):
if target - nums[x] in d:
return d[target-nums[x]],x
else:
d[nums[x]] = x
