[算法分析与设计] leetcode 每周一题: 215. Kth Largest Element in an Array

题目链接：https://leetcode.com/problems/kth-largest-element-in-an-array/description/

题目大意：找出数组中第N大的数字

思路：

1.一开始，我是单纯采用算法概论里面的分治策略解答，将数组每次划分为3类（大于目标值，小于目标值，等于目标值），当递归到N处于等于目标值数组时候，即答案，结果，leetcode中超出内存限制，代码如下：

class Solution {
public:
int findKthLargest(vector<int>& nums, int k) {
vector<int> larger, smaller, same;
int target = nums[0];
for(auto i : nums) {
if(i < target)
smaller.push_back(i);
else if(i == target)
same.push_back(i);
else
larger.push_back(i);
}
if(k <= larger.size()) {
return findKthLargest(larger, k);
}
else if(k <= (larger.size() + same.size()))
return target;
else
return findKthLargest(smaller, k - (larger.size() + same.size()));

}
};后来，网上查了其他博客，发现可以用类似快排的方法解答，代码如下；
class Solution {
public:
int findKthLargest(vector<int>& nums, int k) {
int len = nums.size();
Quick_sort(nums,k-1,0,len-1);
return nums[k-1];
}
void Quick_sort(vector<int>& nums, int k, int left, int right) {
int l = left;
int r = right;
// int len = nums.size();
int key = nums[l];
while(l < r) {
while(r > l && nums[r] < key)
r--;
nums[l] = nums[r];
while(r > l && nums[l] >= key)
l++;
nums[r] = nums[l];

}
nums[l] = key;

if(k < r)
Quick_sort(nums,k,left,r-1);
if(k > r)
Quick_sort(nums,k,r+1,right);

}
};