双指针

1、面试题 02.02. 返回倒数第 k 个节点

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()

2、剑指 Offer 22. 链表中倒数第k个节点

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]

3、面试题 17.11. 单词距离

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

4、面试题 02.04. 分割链表

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

5、面试题 17.21. 直方图的水量

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