LeetCode：721.Accounts Merge賬戶合併（C語言）
阿新 • • 發佈：2021-02-05
技術標籤：LeetCode
題目描述：
給定一個列表 accounts，每個元素 accounts[i] 是一個字串列表，其中第一個元素 accounts[i][0] 是 名稱 (name)，其餘元素是 emails 表示該賬戶的郵箱地址。
現在，我們想合併這些賬戶。如果兩個賬戶都有一些共同的郵箱地址，則兩個賬戶必定屬於同一個人。請注意，即使兩個賬戶具有相同的名稱，它們也可能屬於不同的人，因為人們可能具有相同的名稱。一個人最初可以擁有任意數量的賬戶，但其所有賬戶都具有相同的名稱。
合併賬戶後，按以下格式返回賬戶：每個賬戶的第一個元素是名稱，其餘元素是按順序排列的郵箱地址。賬戶本身可以以任意順序返回。
示例 1：
輸入：
accounts = [[“John”, “[email protected]”, “[email protected]”], [“John”, “[email protected]”], [“John”, “[email protected]”, “[email protected]”], [“Mary”, “[email protected]”]]
輸出：
[[“John”, ‘[email protected]’, ‘[email protected]’, ‘[email protected]’], [“John”, “ [email protected]”], [“Mary”, “[email protected]”]]
解釋：
第一個和第三個 John 是同一個人，因為他們有共同的郵箱地址 “[email protected]”。
第二個 John 和 Mary 是不同的人，因為他們的郵箱地址沒有被其他帳戶使用。
可以以任何順序返回這些列表，例如答案 [[‘Mary’，‘[email protected]’]，[‘John’，‘[email protected]’]，
[‘John’，‘[email protected]’，‘[email protected]
提示：
accounts的長度將在[1，1000]的範圍內。
accounts[i]的長度將在[1，10]的範圍內。
accounts[i][j]的長度將在[1，30]的範圍內。
來源：力扣（LeetCode）
連結：https://leetcode-cn.com/problems/accounts-merge
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解答：
/**
* Return an array of arrays of size *returnSize.
* The sizes of the arrays are returned as *returnColumnSizes array.
* Note: Both returned array and *columnSizes array must be malloced, assume caller calls free().
*/
#define MAXSTRLEN 35
struct HashMap{
char key[MAXSTRLEN];
char name[MAXSTRLEN];
char parent[MAXSTRLEN];
UT_hash_handle hh;
};
struct HashMap *g_hashHead = NULL;
char ***g_res = NULL;
int *g_col = NULL;
int g_resIndex = 0;
#define MAXNUM 10000
void InitMem(int accountsSize)
{
g_hashHead = NULL;
g_res = (char ***)malloc(sizeof(char **) * accountsSize);
for(int i = 0; i < accountsSize; i++) {
g_res[i] = calloc(MAXNUM, sizeof(char*));
}
g_col = calloc(accountsSize, sizeof(int));
g_resIndex = 0;
return;
}
void FreeMem()
{
struct HashMap *s = NULL;
struct HashMap *tmp = NULL;
HASH_ITER(hh,g_hashHead,s,tmp) {
HASH_DEL(g_hashHead,s);
free(s);
}
return;
}
struct HashMap *InsertHash(char *key, char *name, char *parent)
{
struct HashMap *s = NULL;
HASH_FIND_STR(g_hashHead,key,s);
if(s == NULL) {
s = malloc(sizeof(struct HashMap));
strcpy(s->key, key);
strcpy(s->name, name);
strcpy(s->parent, parent);
HASH_ADD_STR(g_hashHead,key,s);
}
return s;
}
struct HashMap *UpadatHash(char *key, char *name, char *parent)
{
struct HashMap *s = NULL;
HASH_FIND_STR(g_hashHead,key,s);
if(s == NULL) {
return NULL;
}
strcpy(s->name, name);
strcpy(s->parent, parent);
return s;
}
struct HashMap *Find(struct HashMap *x )
{
struct HashMap *s = x;
while(strcmp(s->key, s->parent) != 0) {
char parent[MAXSTRLEN] = {0};
strcpy(parent,s->parent);
HASH_FIND_STR(g_hashHead,parent,s);
}
return s;
}
void Union(struct HashMap *a, struct HashMap *b)
{
struct HashMap *aParent = Find(a);
struct HashMap *bParent = Find(b);
//printf("FIND a %s parent %s %s\r\n",a->key,aParent->key,aParent->parent);
//printf("FIND b %s parent %s %s\r\n",b->key,bParent->key,bParent->parent);
if(strcmp(aParent->parent, bParent->parent) != 0) {
(void)UpadatHash(bParent->key, aParent->name, aParent->parent);
(void)UpadatHash(b->key, aParent->name, aParent->parent);
}
return;
}
void ProcessAccount(char *** accounts, int accountsSize, int* accountsColSize)
{
struct HashMap *a = NULL;
struct HashMap *b = NULL;
for (int i = 0; i < accountsSize; i++) {
a = NULL;
b = NULL;
char *name = accounts[i][0];
char *key = accounts[i][0];
char *parent = accounts[i][0];
for(int j = 1; j < accountsColSize[i]; j++) {
key = accounts[i][j];
parent = accounts[i][j];
if(j == 1) {
a = InsertHash(key, name, parent);
continue;
}else{
b = InsertHash(key, name, parent);
}
Union(a,b);
}
}
return;
}
bool IsSameParent(char *aKey, char *bKey)
{
struct HashMap *a = NULL;
struct HashMap *b = NULL;
HASH_FIND_STR(g_hashHead,aKey,a);
HASH_FIND_STR(g_hashHead,bKey,b);
struct HashMap *aParent = Find(a);
struct HashMap *bParent = Find(b);
if(aParent != bParent) {
return false;
}
return true;
}
int SortCmp(const void *a, const void *b)
{
char *aPtr = *(char**)a;
char *bPtr = *(char**)b;
return strcmp(aPtr,bPtr);
}
int GetResIndex(char *key)
{
struct HashMap *s = NULL;
HASH_FIND_STR(g_hashHead,key,s);
for(int i = 0; i < g_resIndex; i++) {
if(g_col[i] > 1) {
if(IsSameParent(s->parent, g_res[i][1])) {
return i;
}
}
}
return -1;
}
void OutPutRes()
{
struct HashMap *s = NULL;
struct HashMap *tmp = NULL;
char parent[MAXSTRLEN] = {0};
HASH_ITER(hh,g_hashHead,s,tmp) {
int resIndex = GetResIndex(s->key);
if(resIndex == -1) {
g_res[g_resIndex][0] = calloc(MAXSTRLEN, sizeof(char));
strcpy(g_res[g_resIndex][0],s->name);
g_res[g_resIndex][1] = calloc(MAXSTRLEN, sizeof(char));
strcpy(g_res[g_resIndex][1],s->key);
g_col[g_resIndex] = 2;
g_resIndex++;
}else{
int index = g_col[resIndex];
g_res[resIndex][index] = calloc(MAXSTRLEN, sizeof(char));
strcpy(g_res[resIndex][index],s->key);
g_col[resIndex]++;
}
}
return;
}
void PrintHash()
{
struct HashMap *s = NULL;
struct HashMap *tmp = NULL;
printf("PrintHash:\r\n");
HASH_ITER(hh,g_hashHead,s,tmp) {
printf("hash: %s %s %s\r\n",s->name,s->key,s->parent);
}
return;
}
void PrintRes()
{
for (int i = 0; i < g_resIndex; i++) {
for(int j = 0; j < g_col[i]; j++) {
printf("%s ",g_res[i][j]);
}
printf("\r\n");
}
return;
}
void SortRes()
{
for(int i = 0; i < g_resIndex; i++) {
qsort(&g_res[i][1], g_col[i] - 1, sizeof(char*), SortCmp);//去掉NAME後 開始排序
}
return;
}
char *** accountsMerge(char *** accounts, int accountsSize, int* accountsColSize, int* returnSize, int** returnColumnSizes){
InitMem(accountsSize);
ProcessAccount(accounts, accountsSize, accountsColSize);
//PrintHash();
OutPutRes();
SortRes();
//PrintRes();
FreeMem();
*returnSize = g_resIndex;
*returnColumnSizes = g_col;
return g_res;
}
執行結果：
Notes：
參考文件：https://leetcode-cn.com/problems/accounts-merge/solution/cyu-yan-bing-cha-ji-hashjie-fa-by-tan-shao-yu-dou-/