diff --git a/problems/面试题02.07.链表相交.md b/problems/面试题02.07.链表相交.md index 1ae01061..0fe37949 100644 --- a/problems/面试题02.07.链表相交.md +++ b/problems/面试题02.07.链表相交.md @@ -13,21 +13,21 @@ 图示两个链表在节点 c1 开始相交: -![](https://code-thinking-1253855093.file.myqcloud.com/pics/20211219221657.png) +![](https://code-thinking-1253855093.file.myqcloud.com/pics/20211219221657.png) 题目数据 保证 整个链式结构中不存在环。 -注意,函数返回结果后,链表必须 保持其原始结构 。 +注意,函数返回结果后,链表必须 保持其原始结构 。 -示例 1: +示例 1: -![](https://code-thinking-1253855093.file.myqcloud.com/pics/20211219221723.png) +![](https://code-thinking-1253855093.file.myqcloud.com/pics/20211219221723.png) 示例 2: -![](https://code-thinking-1253855093.file.myqcloud.com/pics/20211219221749.png) +![](https://code-thinking-1253855093.file.myqcloud.com/pics/20211219221749.png) -示例 3: +示例 3: ![](https://code-thinking-1253855093.file.myqcloud.com/pics/20211219221812.png)![](https://code-thinking-1253855093.file.myqcloud.com/pics/20211219221812.png) @@ -100,7 +100,7 @@ public: ## 其他语言版本 -### Java +### Java ```Java public class Solution { public ListNode getIntersectionNode(ListNode headA, ListNode headB) { @@ -144,11 +144,11 @@ public class Solution { } return null; } - + } ``` -### Python +### Python ```python class Solution: @@ -162,15 +162,15 @@ class Solution: """ cur_a, cur_b = headA, headB # 用两个指针代替a和b - + while cur_a != cur_b: cur_a = cur_a.next if cur_a else headB # 如果a走完了,那么就切换到b走 cur_b = cur_b.next if cur_b else headA # 同理,b走完了就切换到a - + return cur_a ``` -### Go +### Go ```go func getIntersectionNode(headA, headB *ListNode) *ListNode { @@ -208,7 +208,30 @@ func getIntersectionNode(headA, headB *ListNode) *ListNode { } ``` -### javaScript +双指针 + +```go +func getIntersectionNode(headA, headB *ListNode) *ListNode { + l1,l2 := headA, headB + for l1 != l2 { + if l1 != nil { + l1 = l1.Next + } else { + l1 = headB + } + + if l2 != nil { + l2 = l2.Next + } else { + l2 = headA + } + } + + return l1 +} +``` + +### javaScript ```js var getListLen = function(head) { @@ -218,9 +241,9 @@ var getListLen = function(head) { cur = cur.next; } return len; -} +} var getIntersectionNode = function(headA, headB) { - let curA = headA,curB = headB, + let curA = headA,curB = headB, lenA = getListLen(headA), lenB = getListLen(headB); if(lenA < lenB) {