LeetCode面试常考算法——链表

反转链表 LeetCode

输入: 1->2->3->4->5->NULL
输出: 5->4->3->2->1->NULL

【解题关键】以其中一个节点为例，反转其实是将当前节点的next指针指向当前节点的前驱节点。反转链表其实是将所有节点都反转一遍。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode reverseList(ListNode head) {
        ListNode pre = null;
        ListNode next = null;
        while(head != null){
            next = head.next;
            head.next = pre;
            pre = head;
            head = next;
        }
        return pre;
    }
}

环形链表I LeetCode

【解法一】0.5s或者1s不断遍历next，如果可以走到头，也就是说next为null，无环；如果走不到头，有环。

【解法二】使用Set来判重。遍历链表，将每个ListNode都加到Set中，一直遍历到链表尾部，如果Set中存在ListNode，则有环，不存在则无环。

    /**
     * Definition for singly-linked list.
     * class ListNode {
     *     int val;
     *     ListNode next;
     *     ListNode(int x) {
     *         val = x;
     *         next = null;
     *     }
     * }
     */
    public class Solution {
        public boolean hasCycle(ListNode head) {
            Set<ListNode> set = new HashSet<>();
            while (head != null) {
                if (set.contains(head)) {
                    return true;
                }
                set.add(head);
                head = head.next;
            }
            return false;
        }
    }

【解法三】快慢指针法，快指针每次走两步，慢指针每次走一步；两个指针如果可以相遇则有环，不能相遇则无环；

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public boolean hasCycle(ListNode head) {
        /*双指针：快慢指针*/
        if(head == null || head.next == null){
            return false;
        }
        ListNode slow = head.next;
        ListNode fast = head.next.next;
        while(fast != null && fast.next != null){
            slow = slow.next;
            fast = fast.next.next;
            if(slow == fast){
                return true;
            }
        }
        return false;
    }
}

环形链表II LeetCode

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode detectCycle(ListNode head) {
        ListNode slow = head;
        ListNode fast = head;
        while(fast != null && fast.next != null){
            slow = slow.next;
            fast = fast.next.next;
            if(slow == fast){
               ListNode slow2 = head; 
                while (slow2 != slow){
                    slow = slow.next;
                    slow2 = slow2.next;
                }
                return slow;
            }
        }
        return null;
    }
}