adding Problem 23

按照合并两个ListNode的方法，合并多个ListNode。但是效率不高

Author: aQua

Algorithms/0023.merge-k-sorted-lists/README.md

@@ -1,7 +1,7 @@
 # [23. Merge k Sorted Lists](https://leetcode.com/problems/merge-k-sorted-lists/)
 
 ## 题目
-
+Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.
 
 ## 解题思路
 

Algorithms/0023.merge-k-sorted-lists/merge-k-sorted-lists.go

@@ -1,2 +1,70 @@
 package Problem0023
 
+// ListNode 是链接节点
+type ListNode struct {
+	Val  int
+	Next *ListNode
+}
+
+func mergeKLists(lists []*ListNode) *ListNode {
+	if len(lists) == 0 {
+		return nil
+	}
+
+	res := lists[0]
+
+	for i := 1; i < len(lists); i++ {
+		res = merge2Lists(res, lists[i])
+	}
+
+	return res
+}
+
+func merge2Lists(l1 *ListNode, l2 *ListNode) *ListNode {
+	// 有一条链为nil，直接返回另一条链
+	if l1 == nil {
+		return l2
+	}
+	if l2 == nil {
+		return l1
+	}
+
+	// 此时，两条链都不为nil，可以直接使用l.Val，而不用担心panic
+	// 在l1和l2之间，选择较小的节点作为head，并设置好node
+	var head, node *ListNode
+	if l1.Val < l2.Val {
+		head = l1
+		node = l1
+		l1 = l1.Next
+	} else {
+		head = l2
+		node = l2
+		l2 = l2.Next
+	}
+
+	// 循环比较l1和l2的值，始终选择较小的值连上node
+	for l1 != nil && l2 != nil {
+		if l1.Val < l2.Val {
+			node.Next = l1
+			l1 = l1.Next
+		} else {
+			node.Next = l2
+			l2 = l2.Next
+		}
+
+		// 有了这一步，head才是一个完整的链
+		node = node.Next
+	}
+
+	if l1 != nil {
+		// 连上l1剩余的链
+		node.Next = l1
+	}
+
+	if l2 != nil {
+		// 连上l2剩余的链
+		node.Next = l2
+	}
+
+	return head
+}

Algorithms/0023.merge-k-sorted-lists/merge-k-sorted-lists_test.go

@@ -1,8 +1,8 @@
 package Problem0023
 
 import (
-	"testing"
 	"fmt"
+	"testing"
 
 	"github.com/stretchr/testify/assert"
 )
@@ -15,13 +15,13 @@ type question struct {
 // para 是参数
 // one 代表第一个参数
 type para struct {
-	one string
+	one [][]int
 }
 
 // ans 是答案
 // one 代表第一个答案
 type ans struct {
-	one string
+	one []int
 }
 
 func Test_Problem0023(t *testing.T) {
@@ -30,17 +30,72 @@ func Test_Problem0023(t *testing.T) {
 	qs := []question{
 
 		question{
-			para{""},
-			ans{""},
+			para{[][]int{
+				[]int{1, 4, 7},
+				[]int{2, 5, 8},
+				[]int{3, 6, 9},
+			}},
+			ans{[]int{1, 2, 3, 4, 5, 6, 7, 8, 9}},
+		},
+
+		question{
+			para{[][]int{
+				[]int{1, 4, 7},
+				[]int{2, 5, 8},
+				[]int{},
+			}},
+			ans{[]int{1, 2, 4, 5, 7, 8}},
 		},
-	
+
 		// 如需多个测试，可以复制上方元素。
 	}
 
 	for _, q := range qs {
 		a, p := q.ans, q.para
 		fmt.Printf("~~%v~~\n", p)
 
-		ast.Equal(a.one, (p.one), "输入:%v", p)
+		ast.Equal(a.one, l2s(mergeKLists(ss2ls(p.one))), "输入:%v", p)
 	}
 }
+
+// convert *ListNode to []int
+func l2s(head *ListNode) []int {
+	res := []int{}
+
+	for head != nil {
+		res = append(res, head.Val)
+		head = head.Next
+	}
+
+	return res
+}
+
+// convert []int to *ListNode
+func s2l(nums []int) *ListNode {
+	if len(nums) == 0 {
+		return nil
+	}
+
+	res := &ListNode{
+		Val: nums[0],
+	}
+	temp := res
+	for i := 1; i < len(nums); i++ {
+		temp.Next = &ListNode{
+			Val: nums[i],
+		}
+		temp = temp.Next
+	}
+
+	return res
+}
+
+func ss2ls(numss [][]int) []*ListNode {
+	res := []*ListNode{}
+
+	for _, nums := range numss {
+		res = append(res, s2l(nums))
+	}
+
+	return res
+}

kit/ListNode.go

@@ -9,7 +9,7 @@ type ListNode struct {
 
 // convert *ListNode to []int
 func l2s(head *ListNode) []int {
-	var res []int
+	res := []int{}
 
 	for head != nil {
 		res = append(res, head.Val)

kit/ListNode_test.go

@@ -8,7 +8,7 @@ import (
 
 func Test_l2s(t *testing.T) {
 	ast := assert.New(t)
-	ast.Nil(l2s(nil), "输入nil，没有返回nil")
+	ast.Equal([]int{}, l2s(nil), "输入nil，没有返回[]int{}")
 
 	one2three := &ListNode{
 		Val: 1,