LeetCode Day04 19&24&142&02.07

面试题 02.07. 链表相交

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
        ListNode curA = headA;
        ListNode curB = headB;
        int lenA = 0, lenB = 0;
        while(curA != null){
            lenA ++;
            curA = curA.next;
        }

        while(curB != null){
            lenB ++;
            curB = curB.next;
        }
        curA = headA;
        curB = headB;

        //让curA为最长链表的头，lenA为其长度
        if(lenB > lenA){
            //1. 交换lenA lenB
            int tmpLen = lenA;
            lenA = lenB;
            lenB = tmpLen;
            //2. 交换curA curB
            ListNode tmpNode = curA;
            curA = curB;
            curB = tmpNode;
        }

        int gap = lenA - lenB;
        while(gap-- > 0){
            curA = curA.next;
        }

        while(curA != null){
            if(curA == curB){
                return curA;
            }
            curA = curA.next;
            curB = curB.next;
        }

        return null;
    }
}

19. 删除链表的倒数第 N 个结点

class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        ListNode dummyNode = new ListNode(0);
        dummyNode.next = head;
        ListNode fastIndex = dummyNode;
        ListNode slowIndex = dummyNode;

        for(int i = 0; i < n; i ++){
            fastIndex = fastIndex.next;
        }

        while(fastIndex.next != null){
            fastIndex = fastIndex.next;
            slowIndex = slowIndex.next;
        }

        slowIndex.next = slowIndex.next.next;
        return dummyNode.next;
    }
}

24. 两两交换链表中的节点

class Solution {
    public ListNode swapPairs(ListNode head) {
        if(head == null || head.next == null) return head;
        ListNode next = head.next;
        ListNode newNode = swapPairs(next.next);
        next.next = head;
        head.next = newNode;
        return next;
    }
}

142. 环形链表 II

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 index1 = fast;
                ListNode index2 = head;
                while(index1 != index2){
                    index1 = index1.next;
                    index2 = index2.next;
                }

                return index1;
            }
        }

        return null;
    }
}