C语言 | Leetcode C语言题解之第218题天际线问题

题目：

题解：

struct pair {
    int first, second;
};

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

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

bool cmp1(struct pair* a, struct pair* b) {
    return a->second < b->second;
}

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

void push(struct Heap* obj, int x, int y) {
    int p = ++(obj->heapSize), q = p >> 1;
    obj->heap[p] = (struct pair){x, y};
    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;
    }
}

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

bool empty(struct Heap* obj) {
    return obj->heapSize == 0;
}

int cmp(int* a, int* b) {
    return *a - *b;
}

int** getSkyline(int** buildings, int buildingsSize, int* buildingsColSize, int* returnSize, int** returnColumnSizes) {
    int n = buildingsSize;
    struct Heap* heap = malloc(sizeof(struct Heap));
    init(heap, n << 1, cmp1);

    int boundaries[n << 1];
    for (int i = 0; i < n; i++) {
        boundaries[i << 1] = buildings[i][0];
        boundaries[i << 1 | 1] = buildings[i][1];
    }
    qsort(boundaries, n << 1, sizeof(int), cmp);

    int** ret = malloc(sizeof(int*) * (n << 1));
    *returnColumnSizes = malloc(sizeof(int) * (n << 1));
    *returnSize = 0;
    int idx = 0;
    for (int i = 0; i < (n << 1); i++) {
        int boundary = boundaries[i];
        while (idx < n && buildings[idx][0] <= boundary) {
            push(heap, buildings[idx][1], buildings[idx][2]);
            idx++;
        }
        while (!empty(heap) && top(heap).first <= boundary) {
            pop(heap);
        }

        int maxn = empty(heap) ? 0 : top(heap).second;
        if ((*returnSize) == 0 || maxn != ret[(*returnSize) - 1][1]) {
            int* tmp = malloc(sizeof(int) * 2);
            tmp[0] = boundary, tmp[1] = maxn;
            (*returnColumnSizes)[*returnSize] = 2;
            ret[(*returnSize)++] = tmp;
        }
    }
    return ret;
}

原文地址：https://blog.csdn.net/m0_59237910/article/details/140223098

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