LeetCode

206. Reverse Linked List

Reverse a singly linked list.

时间复杂度O(n)，空间复杂度O(1)

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public:    ListNode* reverseList(ListNode* head) {        if (head == NULL || head->next == NULL) return head;        ListNode* tail = NULL;        ListNode* ans = NULL;        while (head) {            tail = head->next;            head->next = ans;            ans = head;            head = tail;        }        return ans;    }};

92. Reverse Linked List II

Reverse a linked list from position m to n. Do it in-place and in one-pass.

For example:
Given 1->2->3->4->5->NULL, m = 2 and n = 4,

return 1->4->3->2->5->NULL.

Note:
Given m, n satisfy the following condition:
1 ? m ? n ? length of list.

反转部分链表，精髓还是反转链表。反转链表需要定义三个结点【敲黑板】，一个是答案，一个是当前结点，一个是前一个结点。时间复杂度O(n)，空间复杂度O(n)

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public:    ListNode* reverseBetween(ListNode* head, int m, int n) {        if (!head) return head;        ListNode node(0);        ListNode* ans = &node;        ans->next = head;        for (int i = 0; i < m - 1; ++i) {            ans = ans->next;        }        ListNode* pre = ans->next;        for (int i = 0; i < n - m; ++i) {            ListNode* cur = pre->next;            pre->next = cur->next;            cur->next = ans->next;            ans->next = cur;        }        return node.next;    }};