LeetCode 题解（120）: Kth Largest Element in an Array

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题目：

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.

题解：

用Quick Sort的divide-and-conquer法，或者用Priority Queue (Max Heap) 数据结构，注意Java和Python都是最小堆，需要转换一下。

C++版解法一：

class Solution {public:int findKthLargest(vector<int>& nums, int k) {return nums[returnXth(nums, 0, nums.size() - 1, k)];}int returnXth(vector<int>& nums, int head, int tail, int k) {int start = head;int end = tail;srand((unsigned)time(0));int pivot = (rand() % (end + 1 - start)) + start;swap(nums[start], nums[pivot]);pivot = start;while (start <= end) {while (start <= end && nums[end] >= nums[pivot]) end--;if (start <= end && nums[end] < nums[pivot]) {swap(nums[pivot], nums[end]);pivot = end;end--;}while (start <= end && nums[start] <= nums[pivot]) start++;if (start <= end && nums[start] > nums[pivot]) {swap(nums[start], nums[pivot]);pivot = start;start++;}}if (pivot > nums.size() - k) {pivot = returnXth(nums, head, pivot - 1, k);}else if (pivot < nums.size() - k) {pivot = returnXth(nums, pivot + 1, tail, k);}return pivot;}};

C++版解法二：

class Solution {public:    int findKthLargest(vector<int>& nums, int k) {        priority_queue<int> a;        for(int i = 0; i < nums.size(); i++) {            a.push(nums[i]);        }                for(int i = 0; i < k - 1; i++)            a.pop();                return a.top();    }};

Java版解法一：

public class Solution {    public int findKthLargest(int[] nums, int k) {        return nums[findXth(nums, 0, nums.length-1, k)];    }        public int findXth(int[] nums, int head, int tail, int k) {        int begin = head;        int end = tail;        int pivot = begin;        int temp;        while(begin <= end) {            while(begin <= end && nums[end] >= nums[pivot]) end--;            if(begin <= end && nums[end] < nums[pivot]) {                temp = nums[pivot];                nums[pivot] = nums[end];                nums[end] = temp;                pivot = end;                end--;            }            while(begin <= end && nums[begin] <= nums[pivot]) begin++;            if(begin <= end && nums[begin] > nums[pivot]) {                temp = nums[pivot];                nums[pivot] = nums[begin];                nums[begin] = temp;                pivot = begin;                begin++;            }        }        if(pivot > nums.length - k)             pivot = findXth(nums, head, pivot-1, k);        else if(pivot < nums.length - k)            pivot = findXth(nums, pivot+1, tail, k);        return pivot;    }}

Java版解法二：

import java.util.PriorityQueue;public class Solution {    public int findKthLargest(int[] nums, int k) {        PriorityQueue<Integer> heap = new PriorityQueue<>();        for(int i = 0; i < nums.length; i++)            heap.add(nums[i]);        for(int i = 0; i < nums.length - k; i++)            heap.poll();        return heap.peek();    }}

Python版解法一：

class Solution:    # @param {integer[]} nums    # @param {integer} k    # @return {integer}    def findKthLargest(self, nums, k):        return nums[self.findXth(nums, 0, len(nums)-1, k)]            def findXth(self, nums, head, tail, k):        begin = head        end = tail        pivot = head        while begin <= end:            while begin <= end and nums[end] >= nums[pivot]:                end -= 1            if begin <= end and nums[end] < nums[pivot]:                nums[end], nums[pivot] = nums[pivot], nums[end]                pivot = end                end -= 1            while begin <= end and nums[begin] <= nums[pivot]:                begin += 1            if begin <= end and nums[begin] > nums[pivot]:                nums[begin], nums[pivot] = nums[pivot], nums[begin]                pivot = begin                begin += 1        if pivot > len(nums) - k:            pivot = self.findXth(nums, head, pivot-1, k)        elif pivot < len(nums) - k:            pivot = self.findXth(nums, pivot+1, tail, k)        return pivot

Python版解法二：

import heapqclass Solution:    # @param {integer[]} nums    # @param {integer} k    # @return {integer}    def findKthLargest(self, nums, k):        data = []        for i in nums:            heapq.heappush(data, -i)        for i in range(k-1):            heapq.heappop(data)        return -heapq.heappop(data)

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