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:
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| Input: [3,2,1,5,6,4] and k = 2
Output: 5
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Example 2:
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| Input: [3,2,3,1,2,4,5,5,6] and k = 4
Output: 4
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Note:
You may assume k is always valid, 1 ≤ k ≤ array’s length.
1 优先队列
维护一个k个元素组成的优先队列(小根堆), 如果遍历的元素i比优先队列里最小的元素大, 就poll掉最小的元素, 再把i加入到队列中.
最后, 优先队列中最小的元素即为第k大的元素.
复杂度 O(n * logk)
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| class Solution {
public int findKthLargest(int[] nums, int k) {
PriorityQueue<Integer> pq = new PriorityQueue<>();
for(int i : nums)
{
if(pq.size() < k)
{
pq.offer(i);
}
else
{
int tmp = pq.peek();
if(i > tmp)
{
pq.poll();
pq.offer(i);
}
}
}
return pq.peek();
}
}
|
2 快速排序的思想
利用快速排序的思想, 随机选取一个哨兵sentinel = nums[begin]
然后将其partition两部分, 比sentinel小的在数组左边, 比sentinel大的在数组右边.
这样, 我们就能求出sentinel在数组中的排名是第end - j + 1大的元素了. j为sentinel经过partition后最终的位置
如果k == sentinel_rank, 那么直接返回sentinel即可.
如果k > sentinel_rank, 那么说明第k大的数在sentinel左边的子数组中, 从nums[begin], 到nums[j - 1]中寻找第k - j大的元素即可
如果k < sentinel_rank, 那么说明第k大的数在sentinel右边的子数组中, 从nums[j+1]到nums[end]中寻找第k大的数即可
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| class Solution {
public int findKthLargest(int[] nums, int k) {
return kth(nums, 0, nums.length - 1, k);
}
private int kth(int[] nums, int begin, int end, int k)
{
int sentinel = nums[begin];
int i = begin;
int j = end;
while(true)
{
while(i < end && nums[i] <= sentinel)
++i;
while(j > begin && nums[j] >= sentinel)
--j;
if(i >= j)
{
int tmp = nums[j];
nums[j] = nums[begin];
nums[begin] = tmp;
break;
}
int tmp = nums[j];
nums[j] = nums[i];
nums[i] = tmp;
}
int sentinel_rank = end - j + 1;
if(sentinel_rank == k)
{
return sentinel;
}
else if(sentinel_rank < k)
{
return kth(nums, begin, j - 1, k - sentinel_rank);
}
else
{
return kth(nums, j + 1, end, k);
}
}
}
|