LeetCode Contains Duplicate

位操作 O(n)

class Solution:
    # @param {integer[]} nums
    # @return {boolean}
    def containsDuplicate(self, nums):
        pos, neg = 0, 0
        for x in nums:
            if x < 0:
                temp = 1 << -x
                if neg & temp != 0:
                    return True
                pos |= temp
            else:
                temp = 1 << x
                if pos & temp != 0:
                    return True
                pos |= temp
        return False

哈希表 O(n)

Python一行

刷题防身 python array leetcode Contains Duplicate Comments(1) 2015年7月16日 15:32