目录

1. 删除链表的倒数第 N 个结点  🌟🌟

2. 最小覆盖子串  🌟🌟🌟

3. 二叉树的层序遍历  🌟🌟

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1. 删除链表的倒数第 N 个结点

给你一个链表，删除链表的倒数第 n 个结点，并且返回链表的头结点。

进阶：你能尝试使用一趟扫描实现吗？

示例 1：

输入：head = [1,2,3,4,5], n = 2
输出：[1,2,3,5]

示例 2：

输入：head = [1], n = 1
输出：[]

示例 3：

输入：head = [1,2], n = 1
输出：[1]

提示：

• 链表中结点的数目为 sz
• 1 <= sz <= 30
• 0 <= Node.val <= 100
• 1 <= n <= sz

出处：

https://edu.csdn.net/practice/26319573

代码：

class ListNode:def __init__(self, val=0, next=None):self.val = valself.next = next
class LinkList:def __init__(self):self.head=Nonedef initList(self, data):self.head = ListNode(data[0])r=self.headp = self.headfor i in data[1:]:node = ListNode(i)p.next = nodep = p.nextreturn rdef convert_list(self,head):ret = []if head == None:returnnode = headwhile node != None:ret.append(node.val)node = node.nextreturn ret
class Solution:def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:v = ListNode(0, head)handle = vindex = []while v is not None:index.append(v)v = v.nextpre = len(index)-n-1next = len(index)-n+1index[pre].next = index[next] if next >= 0 and next < len(index) else Nonereturn handle.next
# %%
l = LinkList()
list1 = [1,2,3,4,5]
head = l.initList(list1)
n = 2
s = Solution()
print(l.convert_list(s.removeNthFromEnd(head, n)))

输出：

[1, 2, 3, 5]

2. 最小覆盖子串

给你一个字符串 s 、一个字符串 t 。返回 s 中涵盖 t 所有字符的最小子串。如果 s 中不存在涵盖 t 所有字符的子串，则返回空字符串 "" 。

注意：如果 s 中存在这样的子串，我们保证它是唯一的答案。

示例 1：

输入：s = "ADOBECODEBANC", t = "ABC"
输出："BANC"

示例 2：

输入：s = "a", t = "a"
输出："a"

提示：

• 1 <= s.length, t.length <= 105
• s 和 t 由英文字母组成

进阶：你能设计一个在 o(n) 时间内解决此问题的算法吗？

以下程序实现了这一功能，请你填补空白处内容：

```python
class Solution(object):
    def minWindow(self, s, t):
        ls_s, ls_t = len(s), len(t)
        need_to_find = [0] * 256
        has_found = [0] * 256
        min_begin, min_end = 0, -1
        min_window = 100000000000000
        for index in range(ls_t):
            need_to_find[ord(t[index])] += 1
        count, begin = 0, 0
        for end in range(ls_s):
            end_index = ord(s[end])
            if need_to_find[end_index] == 0:
                continue
            has_found[end_index] += 1
            if has_found[end_index] <= need_to_find[end_index]:
                count += 1
            if count == ls_t:
                begin_index = ord(s[begin])
                ____________________________;
                if window_ls < min_window:
                    min_begin = begin
                    min_end = end
                    min_window = window_ls
        if count == ls_t:
            return s[min_begin:min_end + 1]
        else:
            return ''
if __name__ == '__main__':
    s = Solution()
    print(s.minWindow('a', 'a'))
```

出处：

https://edu.csdn.net/practice/26319574

代码：

class Solution(object):def minWindow(self, s, t):ls_s, ls_t = len(s), len(t)need_to_find = [0] * 256has_found = [0] * 256min_begin, min_end = 0, -1min_window = 100000000000000for index in range(ls_t):need_to_find[ord(t[index])] += 1count, begin = 0, 0for end in range(ls_s):end_index = ord(s[end])if need_to_find[end_index] == 0:continuehas_found[end_index] += 1if has_found[end_index] <= need_to_find[end_index]:count += 1if count == ls_t:begin_index = ord(s[begin])while need_to_find[begin_index] == 0 or\\has_found[begin_index] > need_to_find[begin_index]:if has_found[begin_index] > need_to_find[begin_index]:has_found[begin_index] -= 1begin += 1begin_index = ord(s[begin])window_ls = end - begin + 1if window_ls < min_window:min_begin = beginmin_end = endmin_window = window_lsif count == ls_t:return s[min_begin:min_end + 1]else:return ''
if __name__ == '__main__':s = Solution()print(s.minWindow(s = "ADOBECODEBANC", t = "ABC"))print(s.minWindow('a', 'a'))

输出：

BANC
a

3. 二叉树的层序遍历

给你一个二叉树，请你返回其按 层序遍历 得到的节点值。 （即逐层地，从左到右访问所有节点）。

示例：
二叉树：[3,9,20,null,null,15,7],

  3/ \\
9  20/  \\15   7

返回其层序遍历结果：

[
[3],
[9,20],
[15,7]
]

出处：

https://edu.csdn.net/practice/26319575

代码：

class TreeNode:def __init__(self, x):self.val = xself.left = Noneself.right = Noneclass Solution(object):def levelOrder(self, root):""":type root: TreeNode:rtype: List[List[int]]"""if not root:return []queue, res = [root], []while queue:size = len(queue)temp = []for i in range(size):data = queue.pop(0)temp.append(data.val)if data.left:queue.append(data.left)if data.right:queue.append(data.right)res.append(temp)return resdef listToTree(lst):if not lst:return Noneroot = TreeNode(lst[0])queue = [root]i = 1while i < len(lst):node = queue.pop(0)if lst[i] is not None:node.left = TreeNode(lst[i])queue.append(node.left)i += 1if i < len(lst) and lst[i] is not None:node.right = TreeNode(lst[i])queue.append(node.right)i += 1return rootif __name__ == '__main__':s = Solution()null = Nonenums = [3,9,20,null,null,15,7]root = listToTree(nums)print(s.levelOrder(root))

输出：

[[3], [9, 20], [15, 7]]

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