kth largest element in an array


Kth Largest Element in an Array
先上最naive的解法,先把数组排序了,然后返回第num.length - k个数值

1
2
3
4
5
6
public int (int[] nums, int k) {
if (nums == null)
return 0;
Arrays.sort(nums);
return nums[nums.length - k];
}

再来用heap做的.维护一个大小为k的最小堆,在遍历过程中,不断将数据压入堆中,如果堆的大小大于k,那么就把堆顶元素拿出来。

1
2
3
4
5
6
7
8
9
10
11
12
13
public int (int[] nums, int k) {
if (nums == null)
return 0;
PriorityQueue<Integer> q = new PriorityQueue<>();
for(int val : nums) {
q.offer(val);

if(q.size() > k) {
q.poll();
}
}
return q.peek();
}

最后再来一中时间复杂度是O(n)的算法。用到了quick sort的思想。

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
public int (int[] nums, int k) {
if (nums == null)
return 0;
int begin = 0, end = nums.length - 1;
while (true) {
int position = partition(nums, begin, end);
if (position == k - 1)
return nums[position];
else if (position < k - 1) {
begin = position + 1;
} else
end = position - 1;
}
}

public int partition(int[] nums, int begin, int end) {
int pivot = nums[begin];
int left = begin + 1, right = end;
while (left <= right) {
while (left <= right && nums[left] >= pivot)
++left;
while (left <= right && nums[right] <= pivot)
--right;
if (left <= right)
swap(nums, left, right);
}
swap(nums, begin, right);
return right;
}

public void swap(int[] nums, int i, int j) {
int temp = nums[i];
nums[i] = nums[j];
nums[j] = temp;
}