力扣703. 数据流中的第 K 大元素-C语言实现-简单题

题目

传送门

设计一个找到数据流中第 k 大元素的类（class）。注意是排序后的第 k 大元素，不是第 k 个不同的元素。

请实现 KthLargest 类：

KthLargest(int k, int[] nums) 使用整数 k 和整数流 nums 初始化对象。
int add(int val) 将 val 插入数据流 nums 后，返回当前数据流中第 k 大的元素。

示例：

输入：

[“KthLargest”, “add”, “add”, “add”, “add”, “add”]
[[3, [4, 5, 8, 2]], [3], [5], [10], [9], [4]]

输出：

[null, 4, 5, 5, 8, 8]

解释：

KthLargest kthLargest = new KthLargest(3, [4, 5, 8, 2]);
kthLargest.add(3); // return 4
kthLargest.add(5); // return 5
kthLargest.add(10); // return 5
kthLargest.add(9); // return 8
kthLargest.add(4); // return 8

提示：

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1 <= k <= 104
0 <= nums.length <= 104
-104 <= nums[i] <= 104
-104 <= val <= 104
最多调用 add 方法 104 次
题目数据保证，在查找第 k 大元素时，数组中至少有 k 个元素

来源：力扣（LeetCode）

解题

模板：

typedef struct {
    
    

} KthLargest;


KthLargest* kthLargestCreate(int k, int* nums, int numsSize) {
    
    

}

int kthLargestAdd(KthLargest* obj, int val) {
    
    

}

void kthLargestFree(KthLargest* obj) {
    
    

}

/**
 * Your KthLargest struct will be instantiated and called as such:
 * KthLargest* obj = kthLargestCreate(k, nums, numsSize);
 * int param_1 = kthLargestAdd(obj, val);
 
 * kthLargestFree(obj);
*/

题解超链接

struct Heap {
    
    
    int* heap;
    int heapSize;
    bool (*cmp)(int, int);
};

void init(struct Heap* obj, int n, bool (*cmp)(int, int)) {
    
    
    obj->heap = malloc(sizeof(int) * (n + 1));
    obj->heapSize = 0;
    obj->cmp = cmp;
}

bool cmp(int a, int b) {
    
    
    return a > b;
}

void swap(int* a, int* b) {
    
    
    int tmp = *a;
    *a = *b, *b = tmp;
}

void push(struct Heap* obj, int x) {
    
    
    int p = ++(obj->heapSize), q = p >> 1;
    obj->heap[p] = x;
    while (q) {
    
    
        if (!obj->cmp(obj->heap[q], obj->heap[p])) {
    
    
            break;
        }
        swap(&(obj->heap[q]), &(obj->heap[p]));
        p = q, q = p >> 1;
    }
}

void pop(struct Heap* obj) {
    
    
    swap(&(obj->heap[1]), &(obj->heap[(obj->heapSize)--]));
    int p = 1, q = p << 1;
    while (q <= obj->heapSize) {
    
    
        if (q + 1 <= obj->heapSize) {
    
    
            if (obj->cmp(obj->heap[q], obj->heap[q + 1])) {
    
    
                q++;
            }
        }
        if (!obj->cmp(obj->heap[p], obj->heap[q])) {
    
    
            break;
        }
        swap(&(obj->heap[q]), &(obj->heap[p]));
        p = q, q = p << 1;
    }
}

int top(struct Heap* obj) {
    
    
    return obj->heap[1];
}

typedef struct {
    
    
    struct Heap* heap;
    int maxSize;
} KthLargest;

KthLargest* kthLargestCreate(int k, int* nums, int numsSize) {
    
    
    KthLargest* ret = malloc(sizeof(KthLargest));
    ret->heap = malloc(sizeof(struct Heap));
    init(ret->heap, k + 1, cmp);
    ret->maxSize = k;
    for (int i = 0; i < numsSize; i++) {
    
    
        kthLargestAdd(ret, nums[i]);
    }
    return ret;
}

int kthLargestAdd(KthLargest* obj, int val) {
    
    
    push(obj->heap, val);
    if (obj->heap->heapSize > obj->maxSize) {
    
    
        pop(obj->heap);
    }
    return top(obj->heap);
}

void kthLargestFree(KthLargest* obj) {
    
    
    free(obj->heap->heap);
    free(obj->heap);
    free(obj);
}