92. Reverse Linked List | & ||

Reverse Linked List I

Reverse a linked list.

Example

For linked list 1->2->3, the reversed linked list is 3->2->1

分析：

典型的3 pointers 问题。

/**
 * Definition for ListNode.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int val) {
 *         this.val = val;
 *         this.next = null;
 *     }
 * }
 */ 
public class Solution {
    public ListNode reverse(ListNode head) {
        if (head == null) return head;
        ListNode prev = null;
        ListNode cur = head;
        ListNode next = null;
        while (cur != null) {
            next = cur.next;
            cur.next = prev;
            prev = cur;
            cur = next;
        }
        return prev;
    }
}

Reverse Linked List II

Reverse a linked list from position m to n.

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

Example

Given 1->2->3->4->5->NULL, m = 2 and n = 4, return 1->4->3->2->5->NULL.

分析：

我们需要把list分成三段。首先得到第一段的最后一个node，然后reverse中间部分。最后把三个部分链接起来。

class Solution {
    public ListNode reverseBetween(ListNode head, int m, int n) {
        if (m == n) return head;
        ListNode tempHead = new ListNode(1);
        tempHead.next = head;

        // reach the end of the first part
        ListNode firstPartEnd = tempHead;
        for (int k = 1; k < m; k++) {
            firstPartEnd = firstPartEnd.next;
        }
        // save it later to connect to the last part.
        ListNode secondPartEnd = firstPartEnd.next;

        // reverse the middle part
        ListNode prev = null;
        ListNode curr = firstPartEnd.next;
        ListNode next = null;

        for (int p = 1; p <= n - m + 1; p++) {
            next = curr.next;
            curr.next = prev;
            prev = curr;
            curr = next;
        }
        // connect three parts
        firstPartEnd.next = prev;
        secondPartEnd.next = current;

        return tempHead.next;
    }
}

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