LeetCode6 Kth Largest Element in an Array

1、题目描述

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.

2、解题思路

题意：从一个无序数组中找出第k个最大值，然后输出

方法A：先进行排序，然后输出下标为k-1的元素。

方法B：使用分治的思想，任意选择一个元素作为基准，大于它的元素存入数组l，小于它的元素存入r，等于它的存入e，如果k不大于|l|，递归调用函数本身，在l中继续寻找第k大的元素；如果|l|

class Solution {
public:
int findKthLargest(vector<int>& nums, int k) {
int left = 0, right = nums.size() - 1;
while (true) {
int pos = partition(nums, left, right);
if (pos == k - 1) return nums[pos];
if (pos > k - 1) right = pos - 1;
else left = pos + 1;
}
}
int partition(vector<int>& nums, int left, int right) {
int pivot = nums[left];
int l = left + 1, r = right;
while (l <= r) {
if (nums[l] < pivot && nums[r] > pivot)
swap(nums[l++], nums[r--]);//exchange of element
if (nums[l] >= pivot) l++;
if (nums[r] <= pivot) r--;
}
swap(nums[left], nums[r]);
return r;
}
};

B：

class Solution {
public:
int findKthLargest(vector<int>& nums, int k) {
int v=nums[0];
vector<int> l,e,r;
int kth;
auto n=nums.size();
for(int i=0;i<n;i++){
int t=nums[i];
if (t>v) l.push_back(t);
else{
if(t==v) e.push_back(t);
else r.push_back(t);
}
}
int n1=l.size();
if (n1>=k) kth=findKthLargest(l,k);
else{
int n2=e.size();
if(n1<k<=n1+n2) kth=v;
else{
kth=findKthLargest(r,k-n1-n2);
}

}
return kth;
}

};