标签:lag 描述 -- tin sort 思路分析 output 转换 out
??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.
Example 1:
Input: [3,2,1,5,6,4] and k = 2
Output: 5
Example 2:
Input: [3,2,3,1,2,4,5,5,6] and k = 4
Output: 4
Note:
You may assume k is always valid, 1 ≤ k ≤ array‘s length.
??题目要求找出未排序数组的第K个最大的数,我们可以转换为找从小到大排列的第n-k个数。我们使用快速排序的partition函数,partition函数选择一个flag值,将数组中小于flag的值放在flag左边,将大于flag的值放在右边,我们要找第n-k个数,可以通过判断flag的位置确定,如果flag的位置正好是n-k,那我们找到答案,否则我们可以将范围缩小继续查找。
public class Solution{
public int findKthLargest(int[] nums,int k){
int n=nums.length-k;
int low=0;
int high=nums.length-1;
int t;
while(low<high){
t=partition(nums,low,high);
if(t>n){
high=t-1;
}else if(t<n){
low=t+1;
}else{
break;
}
}
return nums[n];
}
public int partition(int []nums,int low,int high){
int key=nums[low];
while(low<high){
while(low<high&&nums[high]>=key){
high--;
}
nums[low]=nums[high];
while(low<high&&nums[low]<=key){
low++;
}
nums[high]=nums[low];
}
nums[low]=key;
return low;
}
}
49.Kth Largest Element in an Array
标签:lag 描述 -- tin sort 思路分析 output 转换 out
原文地址:https://www.cnblogs.com/yjxyy/p/11087422.html