给定一个长度为n的数组,返回众数。众数是指数组中出现次数超过n/2次的元素
假设数组非空,众数一定存在
Example 1:
Input: [3,2,3] Output: 3
Example 2:
Input: [2,2,1,1,1,2,2] Output: 2
1:字典,累记数组中出现的各元素的次数,一旦发现超过n/2次的元素就返回该元素
def majorityElement(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
if len(nums)==1:
return nums[0]
numDic = {}
for i in nums:
if numDic.has_key(i):
numDic[i] += 1
if numDic.get(i)>=(len(nums)+1)/2:
return i
else:
numDic[i] = 1
2:利用list.count()方法判断(注意for循环中如果是访问整个nums列表会出现“超出时间限制”的错误)
def majorityElement(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
for i in nums[len(nums)//2:]:
if nums.count(i)>len(nums)//2:
return i
3:sorted(nums)[len(nums)//2]
def majorityElement(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
return sorted(nums)[len(nums)//2]
算法题来自:https://leetcode-cn.com/problems/majority-element/description/