leetcode-220-存在重复元素③*

题目描述：

 

 方法一：二叉搜索树+滑动窗口

方法二：桶排序 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

另：

def containsNearbyAlmostDuplicate(self, nums, k, t):
    if t < 0: return False
    n = len(nums)
    d = {}
    w = t + 1
    for i in range(n):
        m = nums[i] // w
        if m in d:
            return True
        if m - 1 in d and abs(nums[i] - d[m - 1]) < w:
            return True
        if m + 1 in d and abs(nums[i] - d[m + 1]) < w:
            return True
        d[m] = nums[i]
        if i >= k: del d[nums[i - k] // w]
    return False