[LeetCode] 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.

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.

// priority_queue: O(nlogn)复杂度
// 12ms
class Solution {
public:
int findKthLargest(vector<int>& nums, int k) {
priority_queue<int> pq;

for (int i = 0; i < nums.size(); i++)
pq.push(nums[i]);

while (k-- > 1) pq.pop();

return pq.top();
}
};

// 16ms
// MergeSort-like Divide-n-Conque, 时间复杂度O(nlogn)
class Solution {
public:
int findKthLargest(vector<int>& nums, int k) {
const int n = nums.size();

auto lp = findKthLargest(nums, 0, (n - 1) / 2, k);
auto rp = findKthLargest(nums, (n - 1) / 2 + 1, n - 1, k);

auto res = MergeK(lp, rp, k);

return res.back();
}
private:
vector<int> findKthLargest(vector<int>& nums, int start, int end, int k) {
if (start > end)
return {};
else if (start == end)
return {nums[start]};
else {
int mid = start + (end - start) / 2;
auto lp = findKthLargest(nums, start, mid, k);
auto rp = findKthLargest(nums, mid + 1, end, k);

return MergeK(lp, rp, k);
}
}

vector<int> MergeK(vector<int>& lp, vector<int>& rp, int k) {
vector<int> TopK;

int i = 0, j = 0;
while (k > 0 && i < lp.size() && j < rp.size()) {
while (k > 0 && i < lp.size() && j < rp.size() && lp[i] >= rp[j]) {
TopK.push_back(lp[i++]);
k--;
}

while (k > 0 && i < lp.size() && j < rp.size() && lp[i] < rp[j]) {
TopK.push_back(rp[j++]);
k--;
}
}

while (k > 0 && i < lp.size()) {
TopK.push_back(lp[i++]);
k--;
}

while (k > 0 && j < rp.size()) {
TopK.push_back(rp[j++]);
k--;
}

return TopK;
}
};