题目:

Given a constant K and a singly linked list L, you are supposed to reverse the links of every K elements on L. For example, given L being 1→2→3→4→5→6, if K = 3, then you must output 3→2→1→6→5→4; if K = 4, you must output 4→3→2→1→5→6.

Input Specification:

Each input file contains one test case. For each case, the first line contains the address of the first node, a positive N (<= 105) which is the total number of nodes, and a positive K (<=N) which is the length of the sublist to be reversed. The address of a node is a 5-digit nonnegative integer, and NULL is represented by -1.

Then N lines follow, each describes a node in the format:

Address Data Next

where Address is the position of the node, Data is an integer, and Next is the position of the next node.

Output Specification:

For each case, output the resulting ordered linked list. Each node occupies a line, and is printed in the same format as in the input.

Sample Input:

00100 6 4
00000 4 99999
00100 1 12309
68237 6 -1
33218 3 00000
99999 5 68237
12309 2 33218

Sample Output:

00000 4 33218
33218 3 12309
12309 2 00100
00100 1 99999
99999 5 68237
68237 6 -1 分析:主要考查链表逆序。 如果通过数组进行排序,然后每K个逆序输出,对于多余的节点将无法通过测试,所以实现必须使用链表。
代码:
typedef struct reverseNode {

    long address;
int value;
long nextAdd;
struct reverseNode *next; } ReverseNode; ReverseNode *reverseLink(ReverseNode *head, int reverseLen)
{
ReverseNode *new = head->next;
ReverseNode *old = new->next;
int count = ;
while (count < reverseLen) {
ReverseNode *temp = old->next;
old->next = new;
old->nextAdd = new->address;
new = old;
old = temp;
count++;
}
head->next->next = old;
if (!old) {
head->next->nextAdd = -;
} else {
head->next->nextAdd = old->address;
}
ReverseNode *temp = head->next;
head->next = new;
head->nextAdd = new->address;
return temp;
} int main()
{
// 读取输入
long beginAddress;
int number, reverseLen;
scanf("%ld %d %d", &beginAddress, &number, &reverseLen); ReverseNode *head = (ReverseNode *)malloc(sizeof(ReverseNode));
ReverseNode *a[number]; for (int i = ; i < number; i++) {
long address, nextAdd;
int value;
scanf("%ld %d %ld",&address, &value, &nextAdd);
ReverseNode *node = (ReverseNode *)malloc(sizeof(ReverseNode));
node->address = address;
node->value = value;
node->nextAdd = nextAdd;
node->next = ;
a[i] = node;
if (beginAddress == address) {
head->next = node;
}
} // 对输入数据通过链表连接起来
ReverseNode *temp = head->next;
int actualNumber = ;
int recyCount = number;
while (temp->nextAdd != - && recyCount-- > ) {
for (int i = ; i < number; i++) {
ReverseNode *tempNode = a[i];
if (tempNode->address == temp->nextAdd) {
temp->next = tempNode;
temp->nextAdd = tempNode->address;
temp = temp->next;
actualNumber++;
break;
}
}
} // 反转
if (reverseLen > ) {
int reverseCount = actualNumber / reverseLen; // 需要进行反转的次数
ReverseNode *tempHead = head;
while (reverseCount-- > ) {
tempHead = reverseLink(tempHead, reverseLen);
}
} ReverseNode *ptr = head->next;
while (ptr) {
if (ptr->nextAdd == -) {
printf("%.5ld %d -1\n", ptr->address, ptr->value);
} else {
printf("%.5ld %d %.5ld\n", ptr->address, ptr->value, ptr->nextAdd);
}
ptr = ptr->next;
}
}

运行结果:

[测试点5,是使用超级大量数据,由于我在将输入数据通过链表连接起来那一步采用的是嵌套循环,复杂度O(N^2),可能是这里导致的,不确定。 但是总体反转思路是这样的,后面找到原因了再进行更新]

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