Sunday, July 10, 2016

[LeetCode]Intersection of Two Linked Lists | 书影博客


[LeetCode]Intersection of Two Linked Lists | 书影博客
Write a program to find the node at which the intersection of two singly linked lists begins.
For example, the following two linked lists:
A:          a1 → a2
                   ↘
                     c1 → c2 → c3
                   ↗            
B:     b1 → b2 → b3
begin to intersect at node c1.
Note:
  • If the two linked lists have no intersection at all, return null.
  • The linked lists must retain their original structure after the function returns.
  • You may assume there are no cycles anywhere in the entire linked structure.
  • Your code should preferably run in O(n) time and use only O(1) memory.

题目大意:

编程求两个单链表的交点
  • 如果链表没有交点,返回null
  • 链表在函数返回时必须保留原始结构
  • 可以假设链表中没有环
  • 代码最好时间复杂度为O(n),空间复杂度O(1)

解题思路:

官方版:

蛮力法枚举(O(mn) 时间, O(1) 空间)
对于链表A中的每个节点ai,遍历整个链表B,检查B中是否有节点与ai重合
哈希表解法(O(n+m) 时间, O(n) or O(m) 空间)
遍历链表A并将每个节点的地址/引用存储在哈希表中。然后检查链表B中的每个节点bi:如果bi出现在哈希表中,则bi就是交点。
双指针解法 (O(n+m) 时间, O(1) 空间):
维护两个指针pA和pB,初始分别指向A和B。然后让它们分别遍历整个链表,每步一个节点。
当pA到达链表末尾时,让它指向B的头节点(没错,是B);类似的当pB到达链表末尾时,重新指向A的头节点。
如果pA在某一点与pB相遇,则pA/pB就是交点。
下面来看下为什么这个算法可行,考虑两个链表:A = {1,3,5,7,9,11} B = {2,4,9,11},它们的交点是节点'9'。由于B的长度是4 小于 A的长度6,pB会首先到达链表的末尾,由于pB比pA恰好少走2个节点。通过把pB指向A的头,把pA指向B的头,我们现在让pB比pA恰好多走2个节点。所以在第二轮,它们可以保证同时在交点相遇。
如果两个链表有交点,则它们的最后一个节点一定是同一个节点。所以当pA/pB到达链表末尾时,分别记录下A和B的最后一个节点。如果两个链表的末尾节点不一致,说明两个链表没有交点。

本人解题思路:

将链表A的末尾节点endA指向B的头结点headB

交点存在性的判断:

如果此时的链表中存在环路,则说明两个链表相交

交点的确定:

pA指针从headA出发,向前走lenB步(链表B的长度)
将pB指针指向headA,令pA和pB同时向前行进(每次一步),记下它们首次相遇的位置,即为链表的交点

原理:

当pA与pB相遇时,pA恰好比pB领先一个整环的长度(lenB)

http://www.cnblogs.com/yuzhangcmu/p/4128794.html
1. 得到2个链条的长度。
2. 将长的链条向前移动差值(len1 - len2)
3. 两个指针一起前进,遇到相同的即是交点,如果没找到,返回null.
相当直观的解法。空间复杂度O(1), 时间复杂度O(m+n)
 1 public ListNode getIntersectionNode1(ListNode headA, ListNode headB) {
 2         if (headA == null || headB == null) {
 3             return null;
 4         }
 5         
 6         int lenA = getLen(headA);
 7         int lenB = getLen(headB);
 8         
 9         if (lenA > lenB) {
10             while (lenA > lenB) {
11                 headA = headA.next;
12                 lenA--;
13             }
14         } else {
15             while (lenA < lenB) {
16                 headB = headB.next;
17                 lenB--;
18             }
19         }
20         
21         while (headA != null) {
22             if (headA == headB) {
23                 return headA;
24             }
25             headA = headA.next;
26             headB = headB.next;
27         }
28         
29         return null;
30     }
31     
32     public int getLen(ListNode node) {
33         int len = 0;
34         while (node != null) {
35             len++;
36             node = node.next;
37         }
38         return len;
39     }

解完后,打开Leetcode的solution, 找到一个很巧妙的解法。其实与解法1相比应该快不了多少,但是写出来超有B格的。。
Two pointer solution (O(n+m) running time, O(1) memory):
Maintain two pointers pA and pB initialized at the head of A and B, respectively. Then let them both traverse through the lists, one node at a time.
When pA reaches the end of a list, then redirect it to the head of B (yes, B, that's right.); similarly when pB reaches the end of a list, redirect it the head of A.
If at any point pA meets pB, then pA/pB is the intersection node.
To see why the above trick would work, consider the following two lists: A = {1,3,5,7,9,11} and B = {2,4,9,11}, which are intersected at node '9'. Since B.length (=4) < A.length (=6), pB would reach the end of the merged list first, because pB traverses exactly 2 nodes less than pA does. By redirecting pB to head A, and pA to head B, we now ask pB to travel exactly 2 more nodes than pA would. So in the second iteration, they are guaranteed to reach the intersection node at the same time.
If two lists have intersection, then their last nodes must be the same one. So when pA/pB reaches the end of a list, record the last element of A/B respectively. If the two last elements are not the same one, then the two lists have no intersections.
 1 public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
 2         if (headA == null || headB == null) {
 3             return null;
 4         }
 5         
 6         ListNode pA = headA;
 7         ListNode pB = headB;
 8         
 9         ListNode tailA = null;
10         ListNode tailB = null;
11         
12         while (true) {
13             if (pA == null) {
14                 pA = headB;
15             }
16             
17             if (pB == null) {
18                 pB = headA;
19             }
20             
21             if (pA.next == null) {
22                 tailA = pA;
23             }
24             
25             if (pB.next == null) {
26                 tailB = pB;
27             }
28             
29             //The two links have different tails. So just return null;
30             if (tailA != null && tailB != null && tailA != tailB) {
31                 return null;
32             }
33             
34             if (pA == pB) {
35                 return pA;
36             }
37             
38             pA = pA.next;
39             pB = pB.next;
40         }
41     }

http://www.jiuzhang.com/solutions/intersection-of-two-linked-lists/
    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
        if (headA == null || headB == null) {
            return null;
        }
        
        // get the tail of list A.
        ListNode node = headA;
        while (node.next != null) {
            node = node.next;
        }
        node.next = headB;
        ListNode result = listCycleII(headA);
        node.next = null;
        return result;
    }
    
    private ListNode listCycleII(ListNode head) {
        ListNode slow = head, fast = head.next;
        
        while (slow != fast) {
            if (fast == null || fast.next == null) {
                return null;
            }
            
            slow = slow.next;
            fast = fast.next.next;
        }
        
        slow = head;
        fast = fast.next;
        while (slow != fast) {
            slow = slow.next;
            fast = fast.next;
        }
        
        return slow;
    }
http://www.programcreek.com/2014/02/leetcode-intersection-of-two-linked-lists-java/

https://github.com/careercup/CtCI-6th-Edition/blob/master/Java/Ch%2002.%20Linked%20Lists/Q2_07_Intersection/Question.java
public static class Result {
public LinkedListNode tail;
public int size;
public Result(LinkedListNode tail, int size) {
this.tail = tail;
this.size = size;
}
}

public static Result getTailAndSize(LinkedListNode list) {
if (list == null) return null;

int size = 1;
LinkedListNode current = list;
while (current.next != null) {
size++;
current = current.next;
}
return new Result(current, size);
}

public static LinkedListNode getKthNode(LinkedListNode head, int k) {
LinkedListNode current = head;
while (k > 0 && current != null) {
current = current.next;
k--;
}
return current;
}

public static LinkedListNode findIntersection(LinkedListNode list1, LinkedListNode list2) {
if (list1 == null || list2 == null) return null;

/* Get tail and sizes. */
Result result1 = getTailAndSize(list1);
Result result2 = getTailAndSize(list2);

/* If different tail nodes, then there's no intersection. */
if (result1.tail != result2.tail) {
return null;
}

/* Set pointers to the start of each linked list. */
LinkedListNode shorter = result1.size < result2.size ? list1 : list2;
LinkedListNode longer = result1.size < result2.size ? list2 : list1;

/* Advance the pointer for the longer linked list by the difference in lengths. */
longer = getKthNode(longer, Math.abs(result1.size - result2.size));

/* Move both pointers until you have a collision. */
while (shorter != longer) {
shorter = shorter.next;
longer = longer.next;
}

/* Return either one. */
return longer;
}
Read full article from [LeetCode]Intersection of Two Linked Lists | 书影博客

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