查找表_leetcode219

标签：思路   contains   保存   return   add   solution   set   contain   on()   

# 解题思路：用查找表（集合）,保存其K个值的状态  20190302 找工作期间

class Solution(object):
    def containsNearbyDuplicate(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: bool
        """
        if len(nums) <=1 :
            return False

        if k <= 0:
            return False


        record = set()
        for i in range(len(nums)):

            if nums[i] in record:
                return True
            else:
                record.add(nums[i])

            if len(record) == k+1:
                record.remove(nums[i-k])

        return False


# n = [1,0,1,1]
# k = 1

n = [1,2,3,1,2,3]
k = 2
s = Solution()
print s.containsNearbyDuplicate(n,k)

查找表_leetcode219

标签：思路   contains   保存   return   add   solution   set   contain   on()   

原文地址：https://www.cnblogs.com/lux-ace/p/10546925.html