leetcode 684. 冗余连接 ufs

解题思路： ufs 并查集中

关键：一条边的两个节点的父节点不相同则Union，若相同则说明这两个节点之前已经被Union过即此边为多余边。（请细品）
并查集用了路径压缩和按秩归并提高效率。

/**
 * Note: The returned array must be malloced, assume caller calls free().
 */
typedef struct {
    int *p;
    int size;
} ufs_t;
ufs_t * initUFS(int n)
{
    ufs_t *ufs = (ufs_t *) malloc(sizeof(ufs_t));
    ufs->p = (int *) malloc(n * sizeof(int));
    ufs->size = n;
    for (int i = 0; i < n; i++) {
        ufs->p[i] = -1;
    }
    return ufs;
}
void destoryUFS(ufs_t *ufs)
{
    if (ufs != NULL&& ufs->p != NULL) {
        free(ufs->p);
        free(ufs);
    }
}
int findUFS(ufs_t *ufs, int x)
{
    if (ufs->p[x] < 0) {
        return x;
    }

    return ufs->p[x] = findUFS(ufs, ufs->p[x]);
}
int merge(ufs_t *ufs, int x, int y)
{
    int fx = findUFS(ufs, x);
    int fy = findUFS(ufs, y);
    if (fx == fy) {
        return -1;
    }
    ufs->p[fx] += ufs->p[fy];
    ufs->p[fy] = fx;
    return 0;
}

int ufs_set_size(ufs_t *ufs, int x)
{
    int fx = findUFS(ufs, x);
    return abs(ufs->p[fx]);
}
int* findRedundantConnection(int** edges, int edgesSize, int* edgesColSize, int* returnSize)
{
    if (edges == NULL || edgesSize <= 0 || *edgesColSize != 2) {
        *returnSize = 0;
        return NULL;
    }
    ufs_t *ufs = initUFS(edgesSize + 1);
    int i;
    for (i = 0; i < edgesSize; i++) {
        if (merge(ufs, edges[i][0], edges[i][1]) != 0) {
            break;
        }
    }
    destoryUFS(ufs);
    int * res = (int *)malloc(*edgesColSize * sizeof(int));
    res[0] = edges[i][0];
    res[1] = edges[i][1];
    *returnSize = 2;
    return res;
}