网易2016 实习研发工程师 [编程题]寻找第K大 and leetcode 215. Kth Largest Element in an Array

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有一个整数数组，请你根据快速排序的思路，找出数组中第K大的数。

给定一个整数数组a,同时给定它的大小n和要找的K(K在1到n之间)，请返回第K大的数，保证答案存在。

测试样例：

[1,3,5,2,2],5,3

返回：2

note:

注意手写快排的时候：

        while(i < j) {
            while(j > i && a[j] > a[left]) j--;
            while(i < j && a[i] <= a[left]) i++;
            if(i < j) {
                swap(a[i],a[j]);
            }
        }

while(j > i && a[j] > a[left]) j--;
while(i < j && a[i] <= a[left]) i++;

这两个顺序不能反了，不然下标会错位一个

 class Finder {
 public:
     int findKth(vector<int> a, int n, int K) {
         // write code here
         return quickfind(a,,n - ,K);
     }

     int quickfind(vector<int> &a,int left,int right,int K) {
         int i = left,j = right;
         while(i < j) {
             while(j > i && a[j] > a[left]) j--;
             while(i < j && a[i] <= a[left]) i++;
             if(i < j) {
                 swap(a[i],a[j]);
             }
         }
         swap(a[left],a[i]);
         int dis = right - i + ;
         if(dis == K){
             return a[i];
         }
         else if(K < dis) {
             return quickfind(a,i + ,right,K);
         }
         else{
             return quickfind(a,left,i -  ,K - dis);
         }
     }
 };

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leetcode:

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215. Kth Largest Element in an Array

QuestionEditorial Solution

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• Total Accepted: 67793
• Total Submissions: 195182
• Difficulty: Medium

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.

Credits:
Special thanks to @mithmatt for adding this problem and creating all test cases.

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 class Solution {
 public:
     int findKthLargest(vector<int>& nums, int k) {
         return quickfind(nums,,nums.size() - ,k);
     }

     int quickfind(vector<int> &a,int left,int right,int K) {
         int i = left,j = right;
         while(i < j) {
             while(j > i && a[j] > a[left]) j--;
             while(i < j && a[i] <= a[left]) i++;
             if(i < j) {
                 swap(a[i],a[j]);
             }
         }
         swap(a[left],a[i]);
         int dis = right - i + ;
         if(dis == K){
             return a[i];
         }
         else if(K < dis) {
             return quickfind(a,i + ,right,K);
         }
         else{
             return quickfind(a,left,i -  ,K - dis);
         }
     }
 };