[LeetCode]220. 存在重复元素 III
题目链接：https://leetcode-cn.com/problems/contains-duplicate-iii/
题目描述：
给定一个整数数组，判断数组中是否有两个不同的索引 i 和 j，使得 nums [i] 和 nums [j] 的差的绝对值最大为 t，并且 i 和 j 之间的差的绝对值最大为 k。
示例：
示例 1:
输入: nums = [1,2,3,1], k = 3, t = 0
输出: true
示例 2:
输入: nums = [1,0,1,1], k = 1, t = 2
输出: true
示例 3:
输入: nums = [1,5,9,1,5,9], k = 2, t = 3
输出: false
思路：
这道题应该是 Hard 难度的（看通过率就知道了）
简单想法：维护一个长度为 k+1的连续队列， 在这队列里一定任何两个数索引号相差 不会超过k，当队列存在两个数相差为t，那么返回为 true
简单想法代码如下，这个一定要理解，后面只是换了数据结构！
class Solution:
def containsNearbyAlmostDuplicate(self, nums: List[int], k: int, t: int) -> bool:
from collections import deque
import bisect
n = len(nums)
if n <= 1: return False
m = min(n, k + 1)
# 维护一个长度为 k + 1 队列
queue = sorted(nums[:m])
# 要删除数
to_del = deque()
to_del.extendleft(nums[:m])
# 先判断首先 k + 1 队列是否存在满足条件的
for i in range(1, m):
if queue[i] - queue[i - 1] <= t:
return True
for i in range(m, n):
# 移动队列
queue.remove(to_del.pop())
# 二分插入
loc = bisect.bisect_left(queue, nums[i])
queue.insert(loc, nums[i])
# 判断它在队列中左右两边数是否小于等于t
if (loc - 1 >= 0 and nums[i] - queue[loc - 1] <= t) or \
(loc + 1 <= k and queue[loc + 1] - nums[i] <= t):
return True
to_del.appendleft(nums[i])
return False
上面的代码也能过，但是极慢，为什么？
原因就是数组插入删除的时间复杂度为 \(O(n)\)，有没有一种数据结构又能排好序，然而删除添加的时间复杂度有很少呢？\(log(n)\)
当然有了，那就是二叉排序树（BST）Python没有自带的库， Java里有 TreeSet
但是可以自己实现，网上找到的代码，大家感兴趣研究一下，（哭！不想看，后面还有一个桶排序要看一下啊！
from collections import deque
class BSTNode:
def __init__(self, dlnode):
self.ptr = dlnode
self.l = None
self.r = None
class BST:
def __init__(self, head):
self.root = BSTNode(head)
def insert(self, dlnode):
def insertHelper(root, dlnode, min_, max_):
if dlnode.v <= root.ptr.v:
if not root.l:
root.l = BSTNode(dlnode)
if min_:
min_ = min_.ptr
return min_, root.ptr
return insertHelper(root.l, dlnode, min_, root)
else:
if not root.r:
root.r = BSTNode(dlnode)
if max_:
max_ = max_.ptr
return root.ptr, max_
return insertHelper(root.r, dlnode, root, max_)
return insertHelper(self.root, dlnode, None, None)
def delNode(self, node):
tmp, prev = self.root, None
while tmp and node != tmp.ptr:
prev = tmp
if node.v <= tmp.ptr.v:
tmp = tmp.l
else:
tmp = tmp.r
if tmp == None:
return f'Something went wrong, Node {node} not found...'
else:
if tmp.l and tmp.r:
tmp2, prev2 = tmp.r, None
while tmp2.l:
prev2 = tmp2
tmp2 = tmp2.l
if prev2:
prev2.l = tmp2.r
if prev:
if prev.l == tmp:
prev.l = tmp2
elif prev.r == tmp:
prev.r = tmp2
else:
return 'Something went wrong.'
else:
self.root = tmp2
tmp2.l = tmp.l
if tmp2 != tmp.r:
tmp2.r = tmp.r
elif tmp.l:
if prev:
if prev.l == tmp:
prev.l = tmp.l
elif prev.r == tmp:
prev.r = tmp.l
else:
return 'Something went wrong.'
else:
self.root = tmp.l
elif tmp.r:
if prev:
if prev.l == tmp:
prev.l = tmp.r
elif prev.r == tmp:
prev.r = tmp.r
else:
return 'Something went wrong.'
else:
self.root = tmp.r
else:
if prev:
if prev.l == tmp:
prev.l = None
elif prev.r == tmp:
prev.r = None
else:
return 'Something went wrong.'
else:
self.root = tmp.r
del tmp
def popByorder(self):
def inorderPop(root):
nodes = []
if root.l:
nodes += inorderPop(root.l)
nodes += [root.ptr]
if root.r:
nodes += inorderPop(root.r)
return nodes
return inorderPop(self.root)
def __str__(self):
def inorder(root):
s = ''
if root.l:
s += inorder(root.l)
s += f' -> {root.ptr.v}'
if root.r:
s += inorder(root.r)
return s
return 'BST' + inorder(self.root)
class DLNode:
def __init__(self, v):
self.v = v
self.next = None
self.prev = None
class DList:
def __init__(self, v):
self.head = DLNode(v) if isinstance(v, int) else v
self.tail = self.head
def append(self, node):
if self.head == self.tail:
self.head.next = node
self.tail.next = node
node.prev = self.tail
self.tail = node
def insert(self, l, m, r):
if l and r:
l.next = m
m.next = r
r.prev = m
m.prev = l
elif l:
self.tail = m
l.next = m
m.prev = l
elif r:
self.head = m
m.next = r
r.prev = m
def delHead(self):
tmp = self.head
self.head = self.head.next
self.head.prev = None
del tmp
def delTail(self):
tmp = self.tail
self.tail = self.tail.prev
self.tail.next = None
del tmp
def delNode(self, d):
if d == self.head:
self.delHead()
elif d == self.tail:
self.delTail()
else:
d.prev.next = d.next
d.next.prev = d.prev
del d
def __str__(self):
s = f'{self.head.v}'
tmp = self.head.next
while tmp:
s += f' -> {tmp.v}'
tmp = tmp.next
return s
class Solution:
def containsNearbyAlmostDuplicate(self, nums: List[int], k: int, t: int) -> bool:
if not nums or k == 0:
return False
n = len(nums)
m = DLNode(nums[0])
bst = BST(m)
toDel = deque([m])
for x in nums[1:k + 1]:
m = DLNode(x)
_, _ = bst.insert(m)
toDel.append(m)
for i, m in enumerate(bst.popByorder()):
if i == 0:
dl = DList(m)
else:
dl.append(m)
i, j = dl.head, dl.head.next
while j:
if j.v - i.v <= t:
return True
i = i.next
j = j.next
if k < n:
for x in nums[k + 1:]:
d = toDel.popleft()
bst.delNode(d)
m = DLNode(x)
l, r = bst.insert(m)
if (l and m.v - l.v <= t) or (r and r.v - m.v <= t):
return True
toDel.append(m)
return False
是的，还有一种思路，桶排序[1]
大家自行用例子模拟，感觉一下！
时间复杂度：\(O(n)\)
class Solution:
def containsNearbyAlmostDuplicate(self, nums: List[int], k: int, t: int) -> bool:
from collections import OrderedDict
n = len(nums)
if n <= 1 or k < 1 or t < 0: return False
queue = OrderedDict()
for n in nums:
key = n if not t else n // t
for m in [queue.get(key-1), queue.get(key), queue.get(key+1)]:
if m is not None and abs(n - m) <= t:
return True
if len(queue) == k:
queue.popitem(False)
queue[key] = n
return False
