考点：
1、collections.deque(maxlen=k)的使用
2、deque队尾插值append，队头取值的popleft()用法(或者result[0])
class Solution:
def kthToLast(self, head: ListNode, k: int) -> int:
from collections import deque
result = deque(maxlen=k)
result.append(head.val)
while head:
result.append(head.val)
head = head.next
return result.popleft()
考点：和上题一样
class Solution:
def getKthFromEnd(self, head: ListNode, k: int) -> ListNode:
from collections import deque
result = deque(maxlen=k)
while head:
result.append(head)
head = head.next
return result[0]
class Solution:
def findClosest(self, words: List[str], word1: str, word2: str) -> int:
word1_index = None
word2_index = None
global result
result = len(words)
def refresh_result():
global result
if word1_index and word2_index and abs(word1_index - word2_index) < result:
result = abs(word1_index- word2_index)
for idx, word in enumerate(words):
if word == word1:
word1_index = idx
refresh_result()
elif word == word2:
word2_index = idx
refresh_result()
return result
考点：
1、双指针，由前往后遍历，如果碰到比k小的值，和前面指针替换
2、技巧，该题有一个技巧，替换指针即替换元素的值
class Solution:
def partition(self, head: ListNode, x: int) -> ListNode:
left, right = head, head
while right:
if right.val < x:
left.val, right.val = right.val, left.val
left = left.next
right = right.next
return head
考点：
1、该题要脑经急转弯明白，该点水面积为左右最大值得较小者，减去当前点的值，python实现很简单
class Solution:
def trap(self, height: List[int]) -> int:
new_height = [0] + height +[0]
result = 0
for i in range(1, len(new_height)-1):
left_max = max(new_height[:i])
right_max = max(new_height[i+1:])
if new_height[i] < left_max and new_height[i] < right_max:
result += min(left_max, right_max) - new_height[i]
return result
原文：https://www.cnblogs.com/pyclq/p/14094749.html