Palindrome Linked List

Palindrome Linked List

Given a singly linked list, determine if it is a palindrome.

Follow up:
Could you do it in O(n) time and O(1) space?

palindrome: n.迴文（正反讀都一樣的詞語）

題目意思較簡單,我就不翻譯了.

原題鏈接: https://leetcode.com/problems/palindrome-linked-list/

編程想法很簡單,也比較笨.就是遍歷一次鏈表,然後創建一個反向的鏈表,再逐一比較即可.其中有兩次循環,算法時間複雜度應該是O(n).但空間複雜度也是O(n).
如何做到O(1)的空間複雜度還沒有想到.

其中涉及到一個問題, 我印象中, 頭結點不是第一個節點,頭結點的下一個節點,纔是第一個數據節點.所以調試了兩種版本.編程環境是VS,附帶調試代碼.

版本一: 頭結點是第一個數據節點

// Definition for singly-linked list.
struct ListNode {
    int val;
    struct ListNode *next;
};

struct ListNode* createList(int* vals, int size) {
    if ((0 >= size) || (NULL == vals)) {
        return NULL;
    }

    struct ListNode* head = (struct ListNode*) malloc(sizeof(struct ListNode));
    struct ListNode* next = head;
    head->next = NULL;
    head->val  = vals[0];

    for (int i = 1; i < size; ++i) {
        next->next = (struct ListNode*) malloc(sizeof(struct ListNode));
        next->next->val = vals[i];
        next->next->next= NULL;
        next = next->next;
    }
    return head;
}

struct ListNode* reverseList(struct ListNode* head) {
    if (NULL == head) {
        return NULL;
    }

    struct ListNode* revHead = (struct ListNode*) malloc(sizeof(struct ListNode));
    struct ListNode* revNext = NULL;
    struct ListNode* tmp     = head;

    while (NULL != tmp) {
        revHead = (struct ListNode*) malloc(sizeof(struct ListNode));
        revHead->val = tmp->val;
        revHead->next= revNext;
        revNext = revHead;
        tmp = tmp->next;
    }

    return revHead;
}

void freeList(struct ListNode* head) {
    struct ListNode* tmp = head;
    while (NULL != head) {
        tmp = head->next;
        free(head);
        head = tmp;
    }
}

bool isPalindrome(struct ListNode* head) {
    if (NULL == head) {
        return false;
    }

    struct ListNode* revList = reverseList(head);
    struct ListNode* tmp1    = head;
    struct ListNode* tmp2    = revList;
    while (NULL != tmp1->next) {
        tmp1 = tmp1->next;
        tmp2 = tmp2->next;
        if (tmp2->val != tmp1->val) {
            freeList(revList);
            return false;
        }
    }

    freeList(revList);
    return true;
}
int _tmain(int argc, _TCHAR* argv[]) {
    char c = 'Y';
    while (('Y' == c) || ('y' == c)) {

        int size = -1;
        cout << "size: ";
        cin >> size;

        int* Array = (int*) malloc(size * sizeof(int));
        memset(Array, '\0', size * sizeof(int));

        for (int i = 0; i < size; ++i) {
            cout << "The " << i + 1 << "th " << "number: ";
            cin >> Array[i];
        }

        cout << isPalindrome(createList(Array, size)) << endl;

        cout << "Continue(Y/N)? ";
        cin >> c;
    }
    return 0;
}

版本二: 頭結點非第一個數據節點, 就其中兩個函數不一樣

// Definition for singly-linked list.
struct ListNode {
    int val;
    struct ListNode *next;
};

/*
 * 頭結點非第一個節點. */
struct ListNode* createList(int* vals, int size) {
    if ((0 >= size) || (NULL == vals)) {
        return NULL;
    }

    struct ListNode* head = (struct ListNode*) malloc(sizeof(struct ListNode));
    struct ListNode* next = head;

    for (int i = 0; i < size; ++i) {
        next->next = (struct ListNode*) malloc(sizeof(struct ListNode));
        next->next->val = vals[i];
        next->next->next= NULL;
        next = next->next;
    }
    return head;
}

struct ListNode* reverseList(struct ListNode* head) {
    if (NULL == head) {
        return NULL;
    }

    struct ListNode* revHead = (struct ListNode*) malloc(sizeof(struct ListNode));
    struct ListNode* revNext = revHead;
    struct ListNode* tmp     = head;
    revHead->next = NULL;

    while (NULL != tmp->next) {
        tmp = tmp->next;
        revHead = (struct ListNode*) malloc(sizeof(struct ListNode));
        revNext->val = tmp->val;
        revHead->next= revNext;
        revNext = revHead;
    }

    return revHead;
}

void freeList(struct ListNode* head) {
    struct ListNode* tmp = head;
    while (NULL != head) {
        tmp = head->next;
        free(head);
        head = tmp;
    }
}

bool isPalindrome(struct ListNode* head) {
    if (NULL == head) {
        return false;
    }

    struct ListNode* revList = reverseList(head);
    struct ListNode* tmp1    = head;
    struct ListNode* tmp2    = revList;
    while (NULL != tmp1->next) {
        tmp1 = tmp1->next;
        tmp2 = tmp2->next;
        if (tmp2->val != tmp1->val) {
            freeList(revList);
            return false;
        }
    }

    freeList(revList);
    return true;
}

int _tmain(int argc, _TCHAR* argv[]) {
    char c = 'Y';
    while (('Y' == c) || ('y' == c)) {

        int size = -1;
        cout << "size: ";
        cin >> size;

        int* Array = (int*) malloc(size * sizeof(int));
        memset(Array, '\0', size * sizeof(int));

        for (int i = 0; i < size; ++i) {
            cout << "The " << i + 1 << "th " << "number: ";
            cin >> Array[i];
        }

        cout << isPalindrome(createList(Array, size)) << endl;

        cout << "Continue(Y/N)? ";
        cin >> c;
    }
    return 0;
}