Description
Given an integer array nums
and an integer k
, return the kth
largest element in the array.
Note that it is the kth
largest element in the sorted order, not the kth
distinct element.
Example 1:
Input: nums = [3,2,1,5,6,4], k = 2
Output: 5
Example 2:
Input: nums = [3,2,3,1,2,4,5,5,6], k = 4
Output: 4
Constraints:
1 <= k <= nums.length <= 104
104 <= nums[i] <= 104
Solutions
Solution 1
Intuition
templates
void quick_sort(int q[], int l, int r)
{
if (l >= r) return;
int i = l - 1, j = r + 1, x = q[l + r >> 1];
while (i < j)
{
do i ++ ; while (q[i] < x);
do j -- ; while (q[j] > x);
if (i < j) swap(q[i], q[j]);
}
quick_sort(q, l, j), quick_sort(q, j + 1, r);
}
private void quickSort(int[] nums, int l, int r) {
if (l >= r) {
return;
}
int i = l - 1, j = r + 1, x = nums[l + r >> 1];
while (i < j) {
do {
i++;
} while (nums[i] < x);
do {
j--;
} while (nums[j] > x);
if (i < j) {
swap(nums, i, j);
}
}
quickSort(nums, l, j);
quickSort(nums, j + 1, r);
}
private void swap(int[] nums, int i, int j) {
int temp = nums[i];
nums[i] = nums[j];
nums[j] = temp;
}
Code
class Solution {
private int quickSort(int[] nums, int l, int r, int k) {
if (l >= r) {
return nums[l];
}
int i = l - 1, j = r + 1, x = nums[l + r >> 1];
while (i < j) {
do {
i++;
} while (nums[i] < x);
do {
j--;
} while (nums[j] > x);
if (i < j) {
swap(nums, i, j);
}
}
int s1 = j - l + 1 ;
if (k <= s1) {
return quickSort(nums, l, j, k);
} else {
return quickSort(nums, j + 1, r, k - s1);
}
}
private void swap(int[] nums, int i, int j) {
int temp = nums[i];
nums[i] = nums[j];
nums[j] = temp;
}
public int findKthLargest(int[] nums, int k) {
return quickSort(nums, 0, nums.length-1, nums.length - k + 1);
}
}
Complexity
- Time: O(nlogn)
- Space: O(logn)
Solution 2
Intuition
Code
Complexity
- Time:
- Space:
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