hashmap什么样的元素才会跑到链表上

在 HashMap 中，当发生哈希碰撞（即不同的键映射到相同的哈希桶）时，元素会以链表的形式存储在同一个哈希桶中。具体来说，以下情况会导 xroot;
subsets[xroot].rank++;
}
}

void kruskalMST(struct Graph* graph) {
int V = graph->V;
struct Edge result[V];
int e = 0;
int i = 0;

heapSort(graph->edge, graph->E);

struct subset* subsets =
(struct subset*)malloc(V * sizeof(struct subset));

for (int v = 0; v < V; v++) {
subsets[v].parent = v;
subsets[v].rank = 0;
}

while (e < V - 1 && i < graph->E) {
struct Edge next_edge = graph->edge[i++];

int x = find(subsets, next_edge.src);
int y = find(subsets, next_edge.dest);

if (x != y) {
result[e++] = next_edge;
Union(subsets, x, y);
}
}

printf("边集输出：\n");
for (i = 0; i < e; ++i)
printf("%d -- %d == %d\n", result[i].src, result[i].dest,
result[i].weight);
}

最后，我们可以在主函数中生成一个无向图，并调用 Kruskal 算法来求解最小生成树：

int main() {
int V = 4; // 顶点数量
int E = 5; // 边的数量
struct Graph* graph = (struct Graph*)malloc(sizeof(struct Graph));
graph->V = V;
graph->E = E;
graph->edge = (struct Edge*)malloc(graph->E * sizeof(struct Edge));

// 随机生成边集
graph->edge[0].src = 0;
graph->edge[0].dest = 1;
graph->edge[0].weight = 10;

graph->edge[1].src = 0;
graph->edge[1].dest = 2;
致元素存储在链表中：

1. 哈希码相同：不同的键经过哈希函数计算 graph->edge[1].weight = 6;

graph->edge[2].src = 0;
graph->edge[2后得到相同的哈希码。
2. equals() 方法返回 true：当两个键的哈希码相等时].dest = 3;
graph->edge[2].weight = 5;

graph->edge[3].src = 1，它们会被放置在同一个哈希桶中。然后，通过 equals() 方法比较两个键的值。如果 equals() 方法返回 true，则表示这两个键相等，它们会被存储在链表中。
3. 哈希桶中已经存在其他键值对：当哈希桶中已经存在一个或多个键值对时，;
graph->edge[3].dest = 3;
graph->edge[3].weight = 15;

graph->edge新的键值对会被添加到链表的末尾。

需要注意的是，Java 8 中引入了红黑[4].src = 2;
graph->edge[4].dest = 3;
graph->edge[4].weight =树来替代链表，当链表长度超过阈值（默认为8）时，链表会自动转换为红黑树，以提高查找效率。