标签：leetcode
Kth Largest Element in an Array
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.
解法一
void quicksort(vector<int> &nums,int l,int h)
{
int i;
int low=l;
int high=h;
int x=nums[l];
if(l>=h)
return;
while(low<high)
{
while(low<high&&nums[high]<=x)
high--;
if(low<high)
nums[low]=nums[high];
while(low<high&&nums[low]>=x)
low++;
if(low<high)
nums[high]=nums[low];
}
nums[low]=x;
quicksort(nums,l,low-1);
quicksort(nums,low+1,h);
}
int findKthLargest(vector<int>& nums, int k) {
int len=nums.size()-1;
quicksort(nums,0,len);
return nums[k-1];
}
解法二
代码如下：
class Solution {
public:
int findKthLargest(vector<int>& nums, int k) {
int len=nums.size()-1;
return findKth(nums,0,len,k);
}
int findKth(vector<int>&nums,int l,int h,int k)
{
if(l>=h)
return nums[l];
int low=l;
int high=h;
int key=nums[l];
while(low<high)
{
while(low<high&&nums[high]<=key)
high--;
if(low<high)
nums[low]=nums[high];
while(low<high&&nums[low]>=key)
low++;
if(low<high)
nums[high]=nums[low];
}
nums[low]=key;
if(k==low+1) //表示已经 找 到了，直接返回K处的值即可
return nums[low];
else if(k<low+1) //K小于当前枢轴，那么在前面部分找
return findKth(nums,l,low-1,k);
else //K大于当前枢轴，那么在后面部分找
return findKth(nums,low+1,h,k);
}
};
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Kth Largest Element in an Array
标签：leetcode
原文地址：http://blog.csdn.net/yujin753/article/details/48008765
