在未排序的數組中找到第 k 個最大的元素。請注意，你需要找的是數組排序后的第 k 個最大的元素，而不是第 k 個不同的元素。

• 示例 1:

輸入: [3,2,1,5,6,4] 和 k = 2 輸出: 5

• 示例 2:

輸入: [3,2,3,1,2,4,5,5,6] 和 k = 4 輸出: 4

說明:

你可以假設 k 總是有效的，且 1 ≤ k ≤ 數組的長度。

快排參考

#include <iostream> #include <vector> using namespace std;void quickSort(vector<int>& nums, int begin, int end); void swap(int& first, int& second); int getIndex(vector<int>& nums, int start, int end); int main() {vector<int> nums = {1,23,67,24,87,45,25,7,68,58,45,34};quickSort(nums, 0, nums.size() - 1);for (const auto& x: nums) {cout << x << " ";}cout << endl;return 0; }void quickSort(vector<int>& nums, int start, int end) {if (start < end) {int index = getIndex(nums, start, end);quickSort(nums, start, index-1);quickSort(nums, index+1, end);} }int getIndex(vector<int>& nums, int start, int end) {int tmp = nums[start];int low = start;int high = end;while(low < high) {while(low < high && nums[high] >= tmp) {high--;}nums[low] = nums[high];while(low < high && nums[low] <= tmp) {low++;}nums[high] = nums[low];}nums[low] = tmp;return low; }void swap(int& first, int& second) {int tmp = first;first = second;second = tmp; }

• 基于快排解決第K個最大元素

// 這個方法速度奇慢 class Solution { public:int findKthLargest(vector<int>& nums, int k) {int index = getIndex(nums, 0, nums.size()-1);while(index != k-1) {index = index > k-1 ? getIndex(nums, 0, index-1) : getIndex(nums, index+1, nums.size()-1);}return nums[index];}int getIndex(vector<int>& nums, int start, int end) {int low = start;int high = end;int tmp = nums[high];while(low < high) {while(low < high && nums[low] >= tmp) {low++;}nums[high] = nums[low];while(low < high && nums[high] <= tmp) {high--;}nums[low] = nums[high];}nums[low] = tmp;return low;}void swap(int& first, int& second) {int tmp = first;first = second;second = tmp;} }

• 優化，使用隨機的index

class Solution { public:int quickSelect(vector<int>& a, int l, int r, int index) {int q = randomPartition(a, l, r);if (q == index) {return a[q];} else {return q < index ? quickSelect(a, q + 1, r, index) : quickSelect(a, l, q - 1, index);}}inline int randomPartition(vector<int>& a, int l, int r) {int i = rand() % (r - l + 1) + l;swap(a[i], a[r]);return partition(a, l, r);}inline int partition(vector<int>& a, int l, int r) {int x = a[r], i = l - 1;for (int j = l; j < r; ++j) {if (a[j] <= x) {swap(a[++i], a[j]);}}swap(a[i + 1], a[r]);return i + 1;}int findKthLargest(vector<int>& nums, int k) {srand(time(0));return quickSelect(nums, 0, nums.size() - 1, nums.size() - k);} }

總結

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