剑指 Offer II 059. 数据流的第 K 大数值 - 力扣（LeetCode） (leetcode-cn.com)

#define MIN 0x80000000
class KthLargest {
    int* minHeap;
    int K;
public:
    KthLargest(int k, vector<int>& nums) {
        K = k;
        minHeap = new int[k + 1];
        int size = nums.size();
        int ini_scale = min(size, k);
        for (int i = 1; i <= ini_scale; ++i) minHeap[i] = nums[i - 1];
        size <= k ? fill(size + 1, k + 1) : add(nums, k, size);
    }

    int fill(int begin, int end) {
        for (int i = begin; i != end; ++i) minHeap[i] = MIN;
        return effervesce();
    }

    int add(vector<int>& nums, int begin, int end) {
        effervesce();
        for (int i = begin; i != end; ++i) add(nums[i]);
        return 0;
    }

    int add(int val) {
        return val < minHeap[1] ? minHeap[1] : sink(val, 1);
    }

    int effervesce() {
        for (int pos = K / 2; pos; --pos) sink(minHeap[pos], pos);
        return 0;
    }
    //核心代码
    int sink(int pebble, int pos) {
        int bubble;
        while ((bubble = pos * 2) <= K) {
            bubble != K && minHeap[bubble + 1] < minHeap[bubble] ? ++bubble : 0;
            if (pebble <= minHeap[bubble]) break;
            minHeap[pos] = minHeap[bubble];
            pos = bubble;
        }
        minHeap[pos] = pebble;
        return minHeap[1];
    }
};