《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.
");
        } else {
            printf("It's not a palindrome.
");
        }
        
        // print the list
        printf("%d", head->val);
        ptr = head->next;
        while (ptr != nullptr) {
            printf("->%d", ptr->val);
            ptr = ptr->next;
        }
        printf("
");
        
        // delete the list
        while (head != nullptr) {
            ptr = head->next;
            delete head;
            head = ptr;
        }
    }
    
    return 0;
}