マイクロソフト2014筆記試験問題21題解答(少し挫折)
タイトルは上図の通りです.ここでは2つの解法を与える:1つ目は自分が考えている,時間複雑度O(n^2);第2の方法は実験室の大牛の考え,時間複雑度O(n)を実現した.
コードは次のとおりです.
テストを経て、运行は间违いありません!!!
コードは次のとおりです.
void reorder1(NODE * root)
{
int sum = -1, changeTimes, j;
NODE * cur, *next, * stopNode;
NODE * this1;
static int insert_num = 0;
assert(root != NULL);
cur = root;
while (cur)
{
sum++;
cur = cur->next;
}
changeTimes = sum / 2;
if (sum & 0x01)
{
changeTimes +=1;
}
insert_num = changeTimes;
stopNode = root;
while (changeTimes--)
{
stopNode = stopNode->next;
}
while (stopNode->next )
{
this1 = stopNode->next;
stopNode->next = this1->next;
insert_num++;
cur = root;
j = sum - insert_num;
while (j--)
{
cur = cur->next;
}
next = cur->next;
cur->next = this1;
this1->next = next;
}
}
void reorder2(NODE * rootp)
{
NODE * pHead1 = rootp, * pHead2 = rootp, *next1 , * next2, *pre2;
int changeTiems, sum = -1;
while (pHead1)
{
++sum;
pHead1 = pHead1->next;
}
changeTiems = sum / 2;
if (sum & 0x01)
{
changeTiems++;
}
while (changeTiems--)
{
pHead2 = pHead2->next;
}
next2 = pHead2->next;
pHead2->next = NULL;
for (pre2 = NULL; (pHead2 = next2) != NULL;)
{
next2 = pHead2->next;
pHead2->next = pre2;
pre2 = pHead2;
}
next2 = pre2;
pHead1 = rootp;
while ((pHead2 = next2) != NULL)
{
next2 = pHead2->next;
next1 = pHead1->next;
pHead1->next = pHead2;
pHead2->next = next1;
pHead1 = next1;
}
}
テストを経て、运行は间违いありません!!!