Leetcode热题100||150:链表

发布于:2024-08-12 ⋅ 阅读:(131) ⋅ 点赞:(0)

206、反转链表

struct ListNode* reverseList(struct ListNode* head) {
    struct ListNode* prev = NULL;
    struct ListNode* curr = head;
    while (curr) {
        struct ListNode* next = curr->next;
        curr->next = prev;
        prev = curr;
        curr = next;
    }
    return prev;
}

234、回文链表

经典的找中点+反转

struct ListNode* reverseList(struct ListNode* head) {
    struct ListNode* prev = NULL;
    struct ListNode* curr = head;
    while (curr != NULL) {
        struct ListNode* nextTemp = curr->next;
        curr->next = prev;
        prev = curr;
        curr = nextTemp;
    }
    return prev;
}

struct ListNode* endOfFirstHalf(struct ListNode* head) {
    struct ListNode* fast = head;
    struct ListNode* slow = head;
    while (fast->next != NULL && fast->next->next != NULL) {
        fast = fast->next->next;
        slow = slow->next;
    }
    return slow;
}

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

    // 找到前半部分链表的尾节点并反转后半部分链表
    struct ListNode* firstHalfEnd = endOfFirstHalf(head);
    struct ListNode* secondHalfStart = reverseList(firstHalfEnd->next);

    // 判断是否回文
    struct ListNode* p1 = head;
    struct ListNode* p2 = secondHalfStart;
    bool result = true;
    while (result && p2 != NULL) {
        if (p1->val != p2->val) {
            result = false;
        }
        p1 = p1->next;
        p2 = p2->next;
    }

    // 还原链表并返回结果
    firstHalfEnd->next = reverseList(secondHalfStart);
    return result;
}

作者:力扣官方题解
链接:https://leetcode.cn/problems/palindrome-linked-list/solutions/457059/hui-wen-lian-biao-by-leetcode-solution/
来源:力扣(LeetCode)
著作权归作者所有。商业转载请联系作者获得授权,非商业转载请注明出处。

61.旋转链表 

61. 旋转链表icon-default.png?t=N7T8https://leetcode.cn/problems/rotate-list/

给你一个链表的头节点 head ,旋转链表,将链表每个节点向右移动 k 个位置。

分析:把他们连成一个环,再在合适位置切开!

struct ListNode* rotateRight(struct ListNode* head, int k) {
    if (k == 0 || head == NULL || head->next == NULL) {//极端情况的排除——一些不需要翻转的
        return head;
    }
    int n = 1;
    struct ListNode* iter = head;
    while (iter->next != NULL) {
        iter = iter->next;
        n++;
    }
    int add = n - k % n;//简化循环次数
    if (add == n) {
        return head;
    }
    iter->next = head;
    while (add--) {
        iter = iter->next;
    }
    struct ListNode* ret = iter->next;
    iter->next = NULL;
    return ret;
}

**还有一些涉及排序的暂时没做


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