《Cracking the Coding Interview》——第2章：链表——题目7

2014-03-18 02:57

题目：检查链表是否是回文的，即是否中心对称。

解法：我的做法是将链表从中间对半拆成两条，然后把后半条反转，再与前半条对比。对比完了再将后半条反转了拼回去。这样不涉及额外的空间，比反转整条链表然后比较要来的好。如果你反转了整条链表又不用额外空间，接下来跟谁比去呢？

代码：

// 2.7 To Check the given linked list is palindrome or not?
#include <cstdio>
#include <unordered_set>
using namespace std;

struct ListNode {
int val;
struct ListNode *next;
ListNode(int x): val(x), next(nullptr) {};
};

class Solution {
public:
bool isPalindromeList(ListNode *head) {
if (head == nullptr) {
return false;
}
if (head->next == nullptr) {
return true;
}

ListNode *t1, *run, *h2;

t1 = run = head;
while (run->next != nullptr && run->next->next != nullptr) {
t1 = t1->next;
run = run->next->next;
}
h2 = t1->next;
t1->next = nullptr;
h2 = reverseList(h2);

ListNode *p1, *p2;
p1 = head;
p2 = h2;
while (p2 != nullptr) {
if (p1->val != p2->val) {
break;
} else {
p1 = p1->next;
p2 = p2->next;
}
}
bool res = (p2 == nullptr);

h2 = reverseList(h2);
t1->next = h2;

return res;
}
private:
ListNode* reverseList(ListNode* head) {
if (head == nullptr) {
return head;
}

ListNode* new_head = nullptr;
ListNode* ptr;

while (head != nullptr) {
if (new_head == nullptr) {
new_head = head;
head = head->next;
new_head->next = nullptr;
} else {
ptr = head->next;
head->next = new_head;
new_head = head;
head = ptr;
}
}

return new_head;
}
};

int main()
{
int i;
int n;
int val;
struct ListNode *head, *ptr;
Solution sol;

while (scanf("%d", &n) == 1 && n > 0) {
// create a linked list
ptr = head = nullptr;
for (i = 0; i < n; ++i) {
scanf("%d", &val);
if (head == nullptr) {
head = ptr = new ListNode(val);
} else {
ptr->next = new ListNode(val);
ptr = ptr->next;
}
}

// check if the list is a palindrome
if (sol.isPalindromeList(head)) {
printf("It's a palindrome.\n");
} else {
printf("It's not a palindrome.\n");
}

// print the list
printf("%d", head->val);
ptr = head->next;
while (ptr != nullptr) {
printf("->%d", ptr->val);
ptr = ptr->next;
}
printf("\n");

// delete the list
while (head != nullptr) {
ptr = head->next;
delete head;
head = ptr;
}
}

return 0;
}