Leetcode 算法题14

35. Search Insert Position

输入一个排好序的列表和一个目标值，求这个目标值在列表中的索引，不在的话求这个目标值排序时应该在列表中位置的索引

Example 1:

Input: [1,3,5,6], 5
Output: 2

Example 2:

Input: [1,3,5,6], 2
Output: 1

Example 3:

Input: [1,3,5,6], 7
Output: 4

Example 1:

Input: [1,3,5,6], 0
Output: 0

我的代码：

class Solution(object):
def searchInsert(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: int
"""
for i in range(len(nums)):
if nums[i] >= target:
return i
return i+1

大神的代码：一行解决

class Solution(object):
def searchInsert(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: int
"""
return len([x for x in nums if x<target])

53. Maximum Subarray

给出一个列表，求其中连续子列表相加得到的最大数

我的代码：主要思路正确就能解决，一开始没有考虑到全为负数的情况，直接调用max()函数只会增加计算量，在遍历过程中再加了个标志

class Solution(object):
def maxSubArray(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
if len(nums) == 1:return nums[0]
max_ = nums[0]
sum_ = 0
max_num = nums[0]
for i in nums:
sum_ = max(sum_ + i, 0)  #可改进
max_ = max(max_,sum_)
max_num = max(max_num,i)
return max_ if max_num > 0 else max_num

大神的代码：思路大致一致，所设的标志比我更好

def maxSubArray(self, A):
if not A:
return 0

curSum = maxSum = A[0]
for num in A[1:]:
curSum = max(num, curSum + num)  #这步是我没想到的
maxSum = max(maxSum, curSum)

return maxSum

728. Self Dividing Numbers

如果一个数能分别被他包含的每个数整除，则符合要求

输入一个范围，求这个范围内所有符合条件的数列表

Example 1:

Input:
left = 1, right = 22
Output: [1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 15, 22]

我的代码：

class Solution(object):
def selfDividingNumbers(self, left, right):
"""
:type left: int
:type right: int
:rtype: List[int]
"""
def Isdn(n):
for i in str(n):
if i == '0' or n % int(i) != 0:
return False
return True
ans = []
for i in range(left,right+1):
if Isdn(i):
ans.append(i)
return ans

大神的代码：写一个函数判断，然后用filter函数

class Solution(object):
def selfDividingNumbers(self, left, right):
is_self_dividing = lambda num: '0' not in str(num) and all([num % int(digit) == 0 for digit in str(num)])
return filter(is_self_dividing, range(left, right + 1))

As pointed out by @ManuelP, 

[num
% int(digit) == 0 for digit in str(num)]

 creates an entire list which is not necessary. By leaving out the 

[

 and 

]

,
we can make use of generators which are lazy and allows for short-circuit evaluation, i.e. 

all

 will
terminate as soon as one of the digits fail the check.

意思是其实all函数输入一个迭代器就可以了，用[]会计算所有数，而不用[]会产生一个生成器，惰性去计算，一旦不符合要求就会退出

The answer below improves the run time from 128 ms to 95 ms:

class Solution(object):
def selfDividingNumbers(self, left, right):
is_self_dividing = lambda num: '0' not in str(num) and all(num % int(digit) == 0 for digit in str(num))
return filter(is_self_dividing, range(left, right + 1))