From e4787fc37000f438c0a4edcd6cdbfbe21b17b242 Mon Sep 17 00:00:00 2001 From: YDZ Date: Tue, 26 Mar 2019 12:13:30 +0800 Subject: [PATCH] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=20problem=20160?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- .../160. Intersection of Two Linked Lists.go | 38 +++++++++++ .... Intersection of Two Linked Lists_test.go | 68 +++++++++++++++++++ .../README.md | 53 +++++++++++++++ 3 files changed, 159 insertions(+) create mode 100644 Algorithms/160. Intersection of Two Linked Lists/160. Intersection of Two Linked Lists.go create mode 100644 Algorithms/160. Intersection of Two Linked Lists/160. Intersection of Two Linked Lists_test.go create mode 100644 Algorithms/160. Intersection of Two Linked Lists/README.md diff --git a/Algorithms/160. Intersection of Two Linked Lists/160. Intersection of Two Linked Lists.go b/Algorithms/160. Intersection of Two Linked Lists/160. Intersection of Two Linked Lists.go new file mode 100644 index 00000000..ac27e70e --- /dev/null +++ b/Algorithms/160. Intersection of Two Linked Lists/160. Intersection of Two Linked Lists.go @@ -0,0 +1,38 @@ +package leetcode + +import "fmt" + +/** + * Definition for singly-linked list. + * type ListNode struct { + * Val int + * Next *ListNode + * } + */ +func getIntersectionNode(headA, headB *ListNode) *ListNode { + //boundary check + if headA == nil || headB == nil { + return nil + } + + a := headA + b := headB + + //if a & b have different len, then we will stop the loop after second iteration + for a != b { + //for the end of first iteration, we just reset the pointer to the head of another linkedlist + if a == nil { + a = headB + } else { + a = a.Next + } + + if b == nil { + b = headA + } else { + b = b.Next + } + fmt.Printf("a = %v b = %v\n", a, b) + } + return a +} diff --git a/Algorithms/160. Intersection of Two Linked Lists/160. Intersection of Two Linked Lists_test.go b/Algorithms/160. Intersection of Two Linked Lists/160. Intersection of Two Linked Lists_test.go new file mode 100644 index 00000000..95d9a6ec --- /dev/null +++ b/Algorithms/160. Intersection of Two Linked Lists/160. Intersection of Two Linked Lists_test.go @@ -0,0 +1,68 @@ +package leetcode + +import ( + "fmt" + "testing" +) + +type question160 struct { + para160 + ans160 +} + +// para 是参数 +// one 代表第一个参数 +type para160 struct { + one []int + another []int +} + +// ans 是答案 +// one 代表第一个答案 +type ans160 struct { + one []int +} + +func Test_Problem160(t *testing.T) { + + qs := []question160{ + + question160{ + para160{[]int{}, []int{}}, + ans160{[]int{}}, + }, + + question160{ + para160{[]int{3}, []int{1, 2, 3}}, + ans160{[]int{3}}, + }, + + question160{ + para160{[]int{1, 2, 3, 4}, []int{1, 2, 3, 4}}, + ans160{[]int{1, 2, 3, 4}}, + }, + + question160{ + para160{[]int{4, 1, 8, 4, 5}, []int{5, 0, 1, 8, 4, 5}}, + ans160{[]int{8, 4, 5}}, + }, + + question160{ + para160{[]int{1}, []int{9, 9, 9, 9, 9}}, + ans160{[]int{}}, + }, + + question160{ + para160{[]int{0, 9, 1, 2, 4}, []int{3, 2, 4}}, + ans160{[]int{2, 4}}, + }, + } + + fmt.Printf("------------------------Leetcode Problem 160------------------------\n") + + for _, q := range qs { + _, p := q.ans160, q.para160 + fmt.Printf("【input】:%v 【output】:%v\n", p, L2s(getIntersectionNode(S2l(p.one), S2l(p.another)))) + } + fmt.Printf("\n\n\n") +} diff --git a/Algorithms/160. Intersection of Two Linked Lists/README.md b/Algorithms/160. Intersection of Two Linked Lists/README.md new file mode 100644 index 00000000..bd53df67 --- /dev/null +++ b/Algorithms/160. Intersection of Two Linked Lists/README.md @@ -0,0 +1,53 @@ +# [160. Intersection of Two Linked Lists](https://leetcode.com/problems/intersection-of-two-linked-lists/) + +## 题目 + +Write a program to find the node at which the intersection of two singly linked lists begins. + +For example, the following two linked lists: + + +begin to intersect at node c1. + +Example 1: + +```c +Input: intersectVal = 8, listA = [4,1,8,4,5], listB = [5,0,1,8,4,5], skipA = 2, skipB = 3 +Output: Reference of the node with value = 8 +Input Explanation: The intersected node's value is 8 (note that this must not be 0 if the two lists intersect). From the head of A, it reads as [4,1,8,4,5]. From the head of B, it reads as [5,0,1,8,4,5]. There are 2 nodes before the intersected node in A; There are 3 nodes before the intersected node in B. +``` + +Example 2: + +```c +Input: intersectVal = 2, listA = [0,9,1,2,4], listB = [3,2,4], skipA = 3, skipB = 1 +Output: Reference of the node with value = 2 +Input Explanation: The intersected node's value is 2 (note that this must not be 0 if the two lists intersect). From the head of A, it reads as [0,9,1,2,4]. From the head of B, it reads as [3,2,4]. There are 3 nodes before the intersected node in A; There are 1 node before the intersected node in B. +``` + + +Example 3: + +```c +Input: intersectVal = 0, listA = [2,6,4], listB = [1,5], skipA = 3, skipB = 2 +Output: null +Input Explanation: From the head of A, it reads as [2,6,4]. From the head of B, it reads as [1,5]. Since the two lists do not intersect, intersectVal must be 0, while skipA and skipB can be arbitrary values. +Explanation: The two lists do not intersect, so return null. +``` + +Notes: + +- 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 + +找到 2 个链表的交叉点。 + +这道题的思路其实类似链表找环。 + +给定的 2 个链表的长度如果一样长，都从头往后扫即可。如果不一样长，需要先“拼成”一样长。把 B 拼接到 A 后面，把 A 拼接到 B 后面。这样 2 个链表的长度都是 A + B。再依次扫描比较 2 个链表的结点是否相同。 + +