链表 相交结点与环问题

基本问题：

1.两个链表中的第一个公共结点

[解题思路]

a.先求得两个链表的长度，得到链表长度差d

b.根据链表长度差，首先让长链表的指针先走d-1步，之后两个指针一起走，发现相同结点时就是公共结点

    int len1 = 0, len2 = 0;
        ListNode p1 = head1, p2 = head2;
        while (p1 != null) {
            len1++;
            p1 = p1.next;
        }

        while (p2 != null) {
            len2++;
            p2 = p2.next;
        }
        
        ListNode listLong = null, listShort = null;
        int lenDiff = 0; 
        if(len1 > len2){
            listLong = head1;
            listShort = head2;
            lenDiff = len1 - len2;
        } else {
            listLong = head2;
            listShort = head1;
            lenDiff = len2 - len1;
        }
        
        for(int i = 0; i < lenDiff; i++){
            listLong = listLong.next;
        }
        
        while(listLong != null && listShort != null && (listLong != listShort)){
            listLong = listLong.next;
            listShort = listShort.next;
        }

        return listLong;

2.如何判断一个链表中是否存在环

 [解题思路]

和上题类似，维护两个指针，一个指针每次走1步，另一个指针每次走2步，如果快指针追上了慢指针，则链表中存在环

public static boolean checkCircle2(ListNode head) {
        if (head == null) {
            return false;
        }

        ListNode fast = head, slow = head;
        while (fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
            if(slow == fast){
                break;
            }
        }
        return !(fast == null || fast.next == null);
    }

 

扩展问题：

1.求问题2中环的入口点

 如明确知晓链表中存在环，则在环中将链表拆开，则就变成上面的问题1，找到入口结点之后恢复链表

public static ListNode findPortal(ListNode head) {
        if (head == null) {
            return null;
        }

        ListNode fast = head, slow = head;
        ListNode head2 = null, tail = null;
        while (fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
            // meet in circle, break the circle
            if (slow == fast) {
                head2 = fast.next;
                tail = fast;
                fast.next = null;
                break;
            }
        }

        ListNode result = findCommonNode(head, head2);
        tail.next = head2;

        return result;
    }

    private static ListNode findCommonNode(ListNode head1, ListNode head2) {
        if (head1 == null || head2 == null) {
            return null;
        }
        int len1 = 0, len2 = 0;
        ListNode p1 = head1, p2 = head2;
        while (p1 != null) {
            len1++;
            p1 = p1.next;
        }

        while (p2 != null) {
            len2++;
            p2 = p2.next;
        }
        
        ListNode listLong = null, listShort = null;
        int lenDiff = 0; 
        if(len1 > len2){
            listLong = head1;
            listShort = head2;
            lenDiff = len1 - len2;
        } else {
            listLong = head2;
            listShort = head1;
            lenDiff = len2 - len1;
        }
        
        for(int i = 0; i < lenDiff; i++){
            listLong = listLong.next;
        }
        
        while(listLong != null && listShort != null && (listLong != listShort)){
            listLong = listLong.next;
            listShort = listShort.next;
        }

        return listLong;
    }

 这题还有另外的解题思路：http://www.cppblog.com/humanchao/archive/2012/11/12/47357.html

假设两个指针相遇时slow指针走了s步，则fast指针走了2s步(因为慢指针走一步，快指针走两步)

此时fast指针可能绕环走了n圈(n >= 0), 则有2s = s + nr; ==> s = nr;

设链表起点到环的入口点距离为a，环入口点到两指针相遇点距离为x.==>s = a + x;

==>a + x = nr;

==>a + x = (n-1)r + r

==>a + x = (n-1)r + L - a

==>a = (n-1)r + L - x -a;

则我们在发现fast指针和slow指针相遇时，另外使用两个指针，分别指向链表头结点和相遇结点，同时开始走，则他们相遇结点就是

环的入口结点

public static ListNode findPortal2(ListNode head) {
        if (head == null) {
            return null;
        }
        
        ListNode fast = head, slow = head;
        while(fast != null && fast.next != null){
            fast = fast.next.next;
            slow = slow.next;
            if(fast == slow){
                break;
            }
        }
        
        if(fast == null || fast.next == null){
            return null;
        }
        
        ListNode p = head;
        while(p != fast){
            p = p.next;
            fast = fast.next;
        }
        return p;
    }

链表其他问题：

1.求链表倒数第k个结点

2.求链表的中间结点，如链表中结点总数为奇数，返回中间结点；如为偶数，返回中间两个结点的任意一个

问题来源：

http://www.cnblogs.com/sooner/p/3277886.html