添加 problem 160

Algorithms/160. Intersection of Two Linked Lists/160. Intersection of Two Linked Lists.go

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+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
+}

Algorithms/160. Intersection of Two Linked Lists/160. Intersection of Two Linked Lists_test.go

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+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")
+}

Algorithms/160. Intersection of Two Linked Lists/README.md

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+# [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。
+
+