LeetCode C++ 215. Kth Largest Element in an Array【堆/分治】中等

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 .

題意：求出數組中第 k 大的數字，不是第 k 大的獨特的數字，說明重複的數字也會佔據序次。

思路1：排序。

class Solution {
public:
    int findKthLargest(vector<int>& nums, int k) {
        sort(nums.begin(), nums.end(), greater<int>());
        return nums[k - 1];
    }
};

思路2：優先隊列(堆排序)。

class Solution {
public:
    int findKthLargest(vector<int>& nums, int k) {
        priority_queue<int, vector<int>, less<int>> pq;
        for (const auto &i : nums) pq.push(i);
        int ans; 
        for (int i = 0; i < k; ++i) {
            ans = pq.top();
            pq.pop();
        }
        return ans;
    }
};

思路3：利用快速排序的分治策略。如果樞紐元是第 k 個數的話，直接返回，否則遞歸找到第 k 大的數。

class Solution {
public:
    int partition(vector<int>& nums, int l, int r) {        //劃分函數
        if (l >= r) return l;
        int t = nums[l];
        while (l < r) {
            while (l < r && nums[r] < t) --r;
            nums[l] = nums[r];
            while (l < r && nums[l] >= t) ++l; 
            nums[r] = nums[l];
        }
        nums[l] = t;
        return l;
    }
    int select(vector<int>& nums, int l, int r, int k) {
        if (l == r) return nums[l];                          //遞歸邊界
        int p = partition(nums, l, r);                       //劃分後主元的位置
        int m = p - l + 1;                                   //nums[p]在[l,r]中的位置
        if (k == m) return nums[p];                          //說明確定的主元位置正好是k
        else if (k < m) return select(nums, l, p - 1, k);    //第k大的數在左側
        else return select(nums, p + 1, r, k - m);           //第k大的數在右側
    }
    int findKthLargest(vector<int>& nums, int k) {
        int kth = select(nums, 0, nums.size() - 1, k);
        return kth;
    }
};