标签：方法 查看 appear else strong self near leetcode lis
Given an array of size n, find the majority element. The majority element is the element that appears more than ? n/2 ? times.
You may assume that the array is non-empty and the majority element always exist in the array.
这一题可以用排序之后查看序列正中间那个元素的方法来解。但是复杂度是排序的复杂度nlog(n)。 用如下的方法借鉴Moore Voting。
class Solution(object): def majorityElement(self, nums): """ :type nums: List[int] :rtype: int """ major = 0 count = 0 for x in nums: if count == 0: major = x count = 1 elif x == major: count += 1 else: count -= 1 return major
Given an integer array of size n, find all elements that appear more than ? n/3 ? times. The algorithm should run in linear time and in O(1) space.
这一题由于对复杂度的限制是线性的，所以不能用排序。可以使用两个majority的candadite。
class Solution(object): def majorityElement(self, nums): """ :type nums: List[int] :rtype: List[int] """ if not nums: return [] n1, n2, c1, c2 = 0, 1, 0, 0 for num in nums: if num == n1: c1 += 1 elif num == n2: c2 += 1 elif c1 == 0: n1 = num; c1 = 1 elif c2 == 0: n2 = num; c2 = 1 else: c1 -= 1; c2 -= 1 size = len(nums) return [n for n in (n1, n2) if nums.count(n) > size/3]
Leetcode # 169, 229 Majority Element I and II
标签：方法 查看 appear else strong self near leetcode lis
原文地址：http://www.cnblogs.com/lettuan/p/6159520.html
