题目链接
LeetCode 215. 数组中的第K个最大元素[1]
题目描述
在未排序的数组中找到第 k 个最大的元素。请注意，你需要找的是数组排序后的第 k 个最大的元素，而不是第 k 个不同的元素。
示例1
输入：
[3,2,1,5,6,4] 和 k = 2
输出：
5
示例2
输入：
[3,2,3,1,2,4,5,5,6] 和 k = 4
输出：
4
解释：
- 你可以假设 k 总是有效的，且 1 ≤ k ≤ 数组的长度。
的右边区间内，递归寻找即可。否则就在左边区间，递归寻找。
代码
排序（c++）
class Solution {
public:
int findKthLargest(vector<int>& nums, int k) {
sort(nums.begin(), nums.end(), greater<int>());
return nums[k-1];
}
};
小根堆+STL优先队列（c++）
class Solution {
public:
int findKthLargest(vector<int>& nums, int k) {
priority_queue<int, vector<int>, greater<int>> Q;
for (auto x : nums) {
Q.push(x);
while (Q.size() > k) Q.pop();
}
return Q.top();
}
};
大根堆+STL优先队列（c++）
class Solution {
public:
int findKthLargest(vector<int>& nums, int k) {
priority_queue<int> Q;
for (auto x : nums) {
Q.push(x);
while (Q.size() > nums.size()-k+1) Q.pop();
}
return Q.top();
}
};
小根堆+手写（c++）
class Solution {
public:
void adjust(vector<int>& nums, int p, int s) {
while (2*p+1 < s) {
int c1 = 2*p+1;
int c2 = 2*p+2;
int c = (c2<s && nums[c2]<nums[c1]) ? c2 : c1;
if (nums[c] < nums[p]) swap(nums[c], nums[p]);
else break;
p = c;
}
}
int findKthLargest(vector<int>& nums, int k) {
int n = nums.size();
for (int i = k/2-1; i >= 0; --i) {
adjust(nums, i, k);
}
for (int i = k; i < n; ++i) {
if (nums[0] >= nums[i]) continue;
swap(nums[0], nums[i]);
adjust(nums, 0, k);
}
return nums[0];
}
};
大根堆+手写（c++）
class Solution {
public:
void adjust(vector<int>& nums, int p, int s) {
while (2*p+1 < s) {
int c1 = 2*p+1;
int c2 = 2*p+2;
int c = (c2<s && nums[c2]>nums[c1]) ? c2 : c1;
if (nums[c] > nums[p]) swap(nums[c], nums[p]);
else break;
p = c;
}
}
int findKthLargest(vector<int>& nums, int k) {
int n = nums.size();
for (int i = (n-k+1)/2-1; i >= 0; --i) {
adjust(nums, i, (n-k+1));
}
for (int i = (n-k+1); i < n; ++i) {
if (nums[0] <= nums[i]) continue;
swap(nums[0], nums[i]);
adjust(nums, 0, (n-k+1));
}
return nums[0];
}
};
快速选择（c++）
class Solution {
public:
int partition(vector<int>& nums, int l, int r) {
int p = l+rand()%(r-l+1), m = l;
swap(nums[p], nums[r]);
for (int i = l; i < r; ++i) {
if (nums[i] < nums[r]) {
swap(nums[i], nums[m++]);
}
}
swap(nums[m], nums[r]);
return m;
}
int quickSelect(vector<int>& nums, int l, int r, int k) {
if (l == r) return nums[l];
int m = partition(nums, l, r);
if (k == m+1) return nums[m];
if (k < m+1) return quickSelect(nums, l, m-1, k);
return quickSelect(nums, m+1, r, k);
}
int findKthLargest(vector<int>& nums, int k) {
int n = nums.size();
srand((int)time(0));
return quickSelect(nums, 0, n-1, n-k+1);
}
};
排序（python）
class Solution:
def findKthLargest(self, nums: List[int], k: int) -> int:
nums.sort(reverse=True)
return nums[k-1]
小根堆+heapq（python）
class Solution:
def findKthLargest(self, nums: List[int], k: int) -> int:
return heapq.nlargest(k, nums)[-1]
小根堆+手写（python）
class Solution:
def findKthLargest(self, nums: List[int], k: int) -> int:
def adjust(nums, p, s):
while 2*p+1 < s:
c1, c2 = 2*p+1, 2*p+2
c = c2 if (c2<s and nums[c2]<nums[c1]) else c1
if nums[c] < nums[p]:
nums[c], nums[p] = nums[p], nums[c]
else:
break
p = c
n = len(nums)
for i in range(k//2-1, -1, -1):
adjust(nums, i, k)
for i in range(k, n):
if nums[0] >= nums[i]: continue
nums[0], nums[i] = nums[i], nums[0]
adjust(nums, 0, k)
return nums[0]
大根堆+手写（python）
class Solution:
def findKthLargest(self, nums: List[int], k: int) -> int:
def adjust(nums, p, s):
while 2*p+1 < s:
c1, c2 = 2*p+1, 2*p+2
c = c2 if (c2<s and nums[c2]>nums[c1]) else c1
if nums[c] > nums[p]:
nums[c], nums[p] = nums[p], nums[c]
else:
break
p = c
n = len(nums)
for i in range((n-k+1)//2-1, -1, -1):
adjust(nums, i, (n-k+1))
for i in range((n-k+1), n):
if nums[0] <= nums[i]: continue
nums[0], nums[i] = nums[i], nums[0]
adjust(nums, 0, (n-k+1))
return nums[0]
快速选择（python）
import random
class Solution:
def findKthLargest(self, nums: List[int], k: int) -> int:
def partition(nums, l, r):
p, m = l+random.randint(0, r-l), l
nums[p], nums[r] = nums[r], nums[p]
for i in range(l, r):
if nums[i] < nums[r]:
nums[m], nums[i] = nums[i], nums[m]
m += 1
nums[m], nums[r] = nums[r], nums[m]
return m
def quickSelect(nums, l, r, k):
if l == r: return nums[l]
m = partition(nums, l, r)
if k == m+1: return nums[m]
if k < m+1: return quickSelect(nums, l, m-1, k)
return quickSelect(nums, m+1, r, k)
n = len(nums)
return quickSelect(nums, 0, n-1, n-k+1)
参考资料
[1]
LeetCode 面试题 17.09. 第 k 个数: https://leetcode-cn.com/problems/kth-largest-element-in-an-array/
