k:int)->int:# 将问题转化为寻找第n-k个最小元素k=len(nums)-kdefquickSelect(l,r):pivot,p=nums[r],l# 将小于等于pivot的元素移动到左侧foriinrange(l,r):ifnums[i]<=pivot:nums[p],nums[i]=nums[i],nums[p]p+=1# 将pivot放到正确的位置上nums[p],nums[r
链接:http://leetcode.com/problems/kth-largest-element-in-an-array/ 题解: 找数组中第k大的元素,可以用heap,Radix sort或者Quick-Select。不过Quick-Select的Time Compleixty只有在Amorized analysis上才是O(n)。下面是使用Java自带的priority queue min-heap来完成的,算是投机取巧了。二刷要补上Radix sort以...
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. Example 1: Input: [3,2,1,5,6,4] and k = 2 Output: 5 Example 2: Input: [3,2,3,1,2,4,5,5,6] and k = 4 Output: 4 Note: You...
public int findKthLargest(int[] nums, int k) { Arrays.sort(nums); return nums[nums.length - k]; } 解法二 我们没必要把所有数字正确排序,我们可以借鉴快排中分区的思想,这里不细讲了,大家可以去回顾一下快排。 随机选择一个分区点,左边都是大于分区点的数,右边都是小于分区点的数。左部分的个数记...
215. Kth Largest Element in an Array 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. Example 1: Input:[3,2,1,5,6,4]and k = 2 Output: 5...
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. ...
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. 该题可以使用基于快速排序分治思想来解决这个问题。第k大数可以等价为第m=nums.length-...
第k大的数 = 第len-k+1小的数 len=end+1 然后实际是找第k-1索引处的数 , 因此为end+1-k+1-1 (其实这里的定义有点混乱了 , 不管怎样第一次Accepted了 , 之后再改进 ) LeetCode Kth Largest Element in an Array 结果 funcpartition(nums[]int,startint,endint)int{p:=start ...
在上文中,我讲了使用Quickselect在O(n)下解决Kth Largest Element in an Array问题,不过Quickselect和Quicksort一样,在最坏情况下(每次选择pivot的值不巧是最大值或者最小值),时间复杂度会是O(n^2)。这边文章介绍另一种更加稳定的(通常情况下可能没有Quickselect快的)方式,即使用min-heap, 或者也可以叫做...
foriinrange(k,len(nums)): # sinve we are trying to keep the k-largest numbers in the heap # we will add any incoming element that is the larger than the min # of the heap. Because our goal is to have as many large elements ...