暴搜的话,在k大的时候是O(n**2)的复杂度,会超时.

采用一个字典来记录每个value的位置.O(N)

class Solution:
    def containsNearbyDuplicate(self, nums: List[int], k: int) -> bool:
        table = {}
        for i, v in enumerate(nums):
            if table.get(v, -1) == -1:
                table[v] = i
            else:
                if i - table[v] <= k:
                    return True
                else:
                    table[v] = i
        return False