字符串算法——查找数组第K个最大值( Kth Largest Element in an Array)

问题：

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.

思路一：最简单粗暴就是对数组进行排序，然后就可以找到第K大的值，例如采用快速排序

class Solution {
public int findKthLargest(int[] nums, int k) {
//快速排序
int len = nums.length;
dfs(nums,0,len-1);
return nums[len-k];
}
public void dfs(int []nums,int start,int end){
if(start<end){
int i = start;
int j = end;
int key = nums[start];
while(i<j){
while(j>i && nums[j]>key)j--;
if(i<j){
swap(nums,i,j);
}
while(i<j && nums[++i]<key);
if(i<j){
swap(nums,i,j);
}
}
dfs(nums,start,i-1);
dfs(nums,j+1,end);
}
}
public void swap(int[]nums,int i,int j){
int temp = nums[i];
nums[i] = nums[j];
nums[j] = temp;
}
}

思路二：直接采用优先队列的方法，优先队列默认建立小顶堆

class Solution {
public int findKthLargest(int[] nums, int k) {
PriorityQueue<Integer> queue = new PriorityQueue<>();//创建优先队列
for(int i = 0;i<nums.length;i++){
queue.offer(nums[i]);//入队操作
if(queue.size()>k)queue.poll();//判断队列长度是否大于K
}
return queue.poll();//返回第K个最大值
}
}