T1. 合并两个有序链表
改进下：
func Merge(pHead1 *ListNode, pHead2 *ListNode) *ListNode {
newHead := &ListNode{}
cur := newHead
for pHead1 != nil && pHead2 != nil {
if pHead1.Val < pHead2.Val {
cur.Next = pHead1
pHead1 = pHead1.Next
} else { //if else 还是 if continue好用？？？
cur.Next = pHead2
pHead2 = pHead2.Next
}
cur = cur.Next
}
if pHead1 == nil {
cur.Next = pHead2
}
if pHead2 == nil {
cur.Next = pHead1
}
return newHead.Next
}
一个问题：
if else 还是 if continue好用？
旧版：
func Merge(pHead1 *ListNode, pHead2 *ListNode) *ListNode {
if pHead1 == nil && pHead2 == nil {
return pHead1
}
newHead := &ListNode{}
cur := newHead
for pHead1 != nil {
if pHead2 == nil { //改进，if非必要勿放循环里，浪费
cur.Next = pHead1
return newHead.Next
}
if pHead1.Val < pHead2.Val {
cur.Next = pHead1
pHead1 = pHead1.Next
//cur = cur.Next 放在后面
} else {
cur.Next = pHead2
pHead2 = pHead2.Next
//cur = cur.Next
} /*else { //不需要对pHead1 == pHead2 做判断！！！
cur.Next = pHead1
pHead1 = pHead1.Next
cur = cur.Next
cur.Next = pHead2
pHead2 = pHead2.Next
cur = cur.Next
}*/
cur = cur.Next
}
cur.Next = pHead2
return newHead.Next
}
T2. 合并k个有序链表
法一：递归分治实现 法二：新链表和每个链表合并
举例说明复杂度：10个链表合并
法一链表最多遍历三次，至少两次，共24次，法二链表最多遍历9次，遍历次数依次减少，共54次
func mergeKLists(lists []*ListNode) *ListNode {
if lists == nil {
return nil
}
//start := 0
//end := len(lists) - 1
return divide(lists, 0, len(lists)-1)
}
func divide(lists []*ListNode, start, end int) *ListNode {
if start == end {
return lists[start]
}
if start > end {
return nil
}
mid := (start + end) / 2
left := divide(lists, start, mid)
right := divide(lists, mid+1, end)
return merge(left, right)
}
func merge(l1 *ListNode, l2 *ListNode) *ListNode {
newl := &ListNode{}
cur := newl
for l1 != nil && l2 != nil {
if l1.Val < l2.Val {
cur.Next = l1
l1 = l1.Next
//cur = cur.Next
} else {
cur.Next = l2
l2 = l2.Next
}
cur = cur.Next
}
if l1 == nil {
cur.Next = l2
}
if l2 == nil {
cur.Next = l1
}
return newl.Next
}
T3. 链表排序****
归并排序（divide()中的结束条件别忘写）
func sortInList(head *ListNode) *ListNode {
if head == nil || head.Next == nil { // divede()结束条件别忘！！！
return head
}
mid := &ListNode{-1, head}
end := mid
for end != nil && end.Next != nil {
end = end.Next.Next
mid = mid.Next
}
right := mid.Next
mid.Next = nil
l1 := sortInList(head)
l2 := sortInList(right)
return sort(l1, l2)
}
func sort(head1, head2 *ListNode) *ListNode {
new1, new2 := head1, head2
newHead := &ListNode{-1, nil}
res := newHead
for new1 != nil && new2 != nil {
if new1.Val < new2.Val {
newHead.Next = new1
new1 = new1.Next
} else {
newHead.Next = new2
new2 = new2.Next
}
newHead = newHead.Next
}
if new1 != nil {
newHead.Next = new1
}
if new2 != nil {
newHead.Next = new2
}
return res.Next
}
