Find the
kth largest element in an unsorted array. Note that it is the kth largest element in the sorted order, not the kth distinct element.
For example,
Given [3,2,1,5,6,4]
and k = 2, return 5.
Note:
You may assume k is always valid, 1 ≤ k ≤ array's length.
Credits:
Special thanks to @mithmatt for adding this problem and creating all test cases.
我AC的做法:
class Solution:
def findKthLargest(self, nums, k):
"""
:type nums: List[int]
:type k: int
:rtype: int
"""
时间复杂度:O(nlogn) 因为用的是快速排序
66ms,竟然打败了74.14%的python3答案,但我相信这肯定不是这道题的考点
快速选择算法:
复杂度:时间 Avg O(N) Worst O(N^2) 空间 O(1)
class Solution:
def findKthLargest(self, nums, k):
"""
:type nums: List[int]
:type k: int
:rtype: int
"""
pivot = nums[0]
left = []
equal = []
right = []
for i in nums:
if i<pivot:
left.append(i)
elif i==pivot:
equal.append(i)
else:
right.append(i)
if k <= len(right):
return self.findKthLargest(right, k)
elif (k - len(right)) <= len(equal):
return equal[0]
else:
return self.findKthLargest(left, k - len(right) - len(equal))
思想是对的, pivot变成随机选取会比固定选数组第一个值好,但实际AC不了,因为内存溢出
最后:https://segmentfault.com/a/1190000003704825 这篇文章总结的很好