215. Kth Largest Element in an Array(partition逆序排序，index+1 == k)

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.

 

class Solution {
public:
    /**
     * @param n: An integer
     * @param nums: An array
     * @return: the Kth largest element
     */
    //第k大的数用快排: 4 5 3 2 5
    int partition(vector<int>& nums,int left,int right) {
        int pivot = nums[right], k=left;
        if(left < right){
            for(int i=left;i<right;i++){
                //逆序
                if(nums[i] >= pivot){
                    swap(nums[k++],nums[i]);
                }
            }
            swap(nums[k],nums[right]);
            return k;
        }
    }
    int search(vector<int> &nums,int n,int left,int right) {
        int index = partition(nums,left,right);
        if(index+1-left == n) return nums[index];
        else if(index+1-left > n) return search(nums,n,left,index-1);
　　　　  //右半部分：只找n-(index+1-left)个元素
        else return search(nums,n-(index+1-left),index+1,right);
    }
    int kthLargestElement(int n, vector<int> &nums) {
        int left =0,right=nums.size()-1;
        return search(nums,n,left,right);
    }
};

 

class Solution {
public:
    //关键 逆序partition,从大到小  3 2 1 5 6 4 
    int partition(vector<int>& nums,int left,int right) {
        int pivot = nums[right],k = left;
        if(left < right){
            for(int i=left;i < right;i++){
                if(nums[i] >= pivot){
                    swap(nums[i],nums[k++]);
                }
            }
            //只能到right-1,否则k有问题。
            swap(nums[k],nums[right]);
        }
        return k;
    }
    
    int Search(vector<int>& nums,int k) {
        int left = 0,right = nums.size()-1;
        int index = partition(nums,left,right);
        if(index+1 == k) return nums[index];
        while(index+1 != k){
            //第k大的数在右半边
            if(index+1 < k){
                left = index+1;
                index = partition(nums,left,right);
            }else{
                right = index -1;
                index = partition(nums,left,right);
            }
        }
        return nums[index];
    }
    
    int findKthLargest(vector<int>& nums, int k) {
        return Search(nums,k);
    }
};

 

class Solution {
public:
    int findKthLargest(vector<int>& nums, int k) {
        return quickSelect(nums, k);
    }
    
private:
    int quickSelect(vector<int>& nums, int k) {
        // 随机选择基准数
        int pivot = nums[rand() % nums.size()];
        // 将大于、小于、等于 pivot 的元素划分至 big, small, equal 中
        vector<int> big, equal, small;
        for (int num : nums) {
            if (num > pivot)
                big.push_back(num);
            else if (num < pivot)
                small.push_back(num);
            else
                equal.push_back(num);
        }
        // 第 k 大元素在 big 中，递归划分
        if (k <= big.size())
            return quickSelect(big, k);
        // 第 k 大元素在 small 中，递归划分
        if (nums.size() - small.size() < k)
            return quickSelect(small, k - nums.size() + small.size());
        // 第 k 大元素在 equal 中，直接返回 pivot
        return pivot;
    }
};