## 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：

• 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：

### 原理：

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.

``` 1 public ListNode getIntersectionNode1(ListNode headA, ListNode headB) {
3             return null;
4         }
5
8
9         if (lenA > lenB) {
10             while (lenA > lenB) {
12                 lenA--;
13             }
14         } else {
15             while (lenA < lenB) {
17                 lenB--;
18             }
19         }
20
21         while (headA != null) {
24             }
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     }```

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) {
3             return null;
4         }
5
8
9         ListNode tailA = null;
10         ListNode tailB = null;
11
12         while (true) {
13             if (pA == null) {
15             }
16
17             if (pB == null) {
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     }```

```    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
return null;
}

// get the tail of list A.
while (node.next != null) {
node = node.next;
}
node.next = null;
return result;
}

while (slow != fast) {
if (fast == null || fast.next == null) {
return null;
}

slow = slow.next;
fast = fast.next.next;
}

fast = fast.next;
while (slow != fast) {
slow = slow.next;
fast = fast.next;
}

return slow;
}```

public static class Result {
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;
while (current.next != null) {
size++;
current = current.next;
}
return new Result(current, size);
}

while (k > 0 && current != null) {
current = current.next;
k--;
}
return current;
}

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;
}