题目描述:
将两个有序链表合并为一个新的有序链表并返回。新链表是通过拼接给定的两个链表的所有节点组成的。
示例:
输入:1->2->4, 1->3->4
输出:1->1->2->3->4->4
Go语言版本解决方案:
package merge_two_sorted_lists
// ListNode 是链接节点
type ListNode struct {
Val int
Next *ListNode
}
func mergeTwoLists(l1 *ListNode, l2 *ListNode) *ListNode {
//如果有一条链为nil,直接返回另一条链
if l1 == nil {
return l2
}
if l2 == nil {
return l1
}
//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
}
//至此,head才是一个完整的链
node = node.Next
}
//剩余节点的处理
if l1 != nil {
node.Next = l1
}
if l2 != nil {
node.Next = l2
}
return head
}
package merge_two_sorted_lists
import (
"fmt"
"github.com/stretchr/testify/assert"
"testing"
)
type question struct {
para
ans
}
//para 是参数
//one 代表第一个参数
type para struct {
one []int
two []int
}
// ans 代表答案
// one 代表第一个答案
type ans struct {
one []int
}
func TestMergeTwoSortedLists(t *testing.T) {
ast := assert.New(t)
qs := []question{
question{
para{[]int{}, []int{1, 3, 5, 7}},
ans{[]int{1, 3, 5, 7}},
},
question{
para{[]int{1, 3, 5, 7}, []int{}},
ans{[]int{1, 3, 5, 7}},
},
question{
para{[]int{1, 3, 5, 7}, []int{2, 4, 6, 8}},
ans{[]int{1, 2, 3, 4, 5, 6, 7, 8}},
},
question{
para{[]int{10, 20, 30}, []int{1, 2, 3}},
ans{[]int{1, 2, 3, 10, 20, 30}},
},
question{
para{[]int{1, 3, 5},
[]int{2, 4, 6, 8, 10},
},
ans{[]int{1, 2, 3, 4, 5, 6, 8, 10}},
},
question{
para{[]int{1, 3, 5, 7, 9},
[]int{2, 4, 6},
},
ans{[]int{1, 2, 3, 4, 5, 6, 7, 9}},
},
}
for _, q := range qs {
a, p := q.ans, q.para
fmt.Printf("~~%v~~\n", p)
ast.Equal(a.one, l2s(mergeTwoLists(s2l(p.one), s2l(p.two))), "输入:%v", p)
}
}
// convert *ListNode to []int
func l2s(head *ListNode) []int {
var 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
}