finish Problem 25

Author: aQua

Algorithms/0025.reverse-nodes-in-k-group/README.md

@@ -17,7 +17,9 @@ For k = 2, you should return: 2->1->4->3->5
 For k = 3, you should return: 3->2->1->4->5
 
 ## 解题思路
+题目要求,把一条链上的每k个节点进行逆转，不足k个的末尾，则不需要逆转。
 
+详见注释
 
 ## 总结
 

Algorithms/0025.reverse-nodes-in-k-group/reverse-nodes-in-k-group.go

@@ -7,34 +7,63 @@ package Problem0025
  *     Next *ListNode
  * }
  */
+
 func reverseKGroup(head *ListNode, k int) *ListNode {
+	if head == nil || head.Next == nil || k < 2 {
+		return head
+	}
+
+	next, ok := needReverse(head, k)
+	if ok {
+		head, tail := reverse(head)
+		// 递归
+		// 把整理好了的前k个节点的尾部，指向整理好了的后面节点的head
+		tail.Next = reverseKGroup(next, k)
+		return head
+	}
 
-	return nil
+	return head
 }
-func canReverse(head *ListNode, k int) (*ListNode, bool) {
+
+// 判断是否有前k个节点需要逆转。
+// 需要的话
+// 会把KthNode.Next = nil，把k和k+1节点斩断，便于前k个节点的逆转。
+func needReverse(head *ListNode, k int) (begin *ListNode, ok bool) {
 	for head != nil {
-		k--
-		if k == 0 {
-			return head.Next, true
+		if k == 1 {
+			begin = head.Next
+			// 把前k与后面的节点斩断, 便于reverse
+			head.Next = nil
+			return begin, true
 		}
+
+		head = head.Next
+		k--
 	}
+
 	return nil, false
 }
 
+// 返回逆转后的首尾节点
 func reverse(head *ListNode) (first, last *ListNode) {
 	if head == nil || head.Next == nil {
 		return head, nil
 	}
-	if head.Next.Next == nil {
-		last = head
-		first = head.Next
-		first.Next = last
+
+	gotLast := false
+
+	for head != nil {
+		temp := head.Next
+		head.Next = first
+		first = head
+		head = temp
+
+		if !gotLast {
+			last = first
+			gotLast = true
+		}
 	}
-	var temp *ListNode
-	last = head
-	first, temp = reverse(head.Next)
-	temp.Next = last
-	//last.Next = nil
+
 	return first, last
 }
 

Algorithms/0025.reverse-nodes-in-k-group/reverse-nodes-in-k-group_test.go

@@ -38,6 +38,14 @@ func Test_Problem0025(t *testing.T) {
 			ans{[]int{3, 2, 1, 4, 5}},
 		},
 
+		question{
+			para{
+				[]int{1, 2, 3, 4, 5},
+				1,
+			},
+			ans{[]int{1, 2, 3, 4, 5}},
+		},
+
 		// 如需多个测试，可以复制上方元素。
 	}
 
@@ -48,10 +56,20 @@ func Test_Problem0025(t *testing.T) {
 		ast.Equal(a.one, l2s(reverseKGroup(s2l(p.one), p.two)), "输入:%v", p)
 	}
 }
-
+func Test_needReverse(t *testing.T) {
+	head := s2l([]int{1, 2, 3, 4, 5, 6})
+	begin, ok := needReverse(head, 4)
+	assert.True(t, ok, "长度足够的链却提示不能逆转")
+	assert.Equal(t, []int{1, 2, 3, 4}, l2s(head), "前链不对, 没有被斩断")
+	assert.Equal(t, []int{5, 6}, l2s(begin), "后链不对")
+}
 func Test_reverse(t *testing.T) {
-	head, _ := reverse(s2l([]int{1, 2, 3}))
-	assert.Equal(t, []int{3, 2, 1}, l2s(head), "无法逆转链")
+	first, last := reverse(s2l([]int{1, 2, 3}))
+	assert.Equal(t, []int{3, 2, 1}, l2s(first), "无法逆转链")
+	assert.Equal(t, 1, last.Val, "链尾部的值不对")
+
+	Nil, _ := reverse(s2l([]int{}))
+	assert.Nil(t, Nil, "无法逆转空链")
 }
 
 // convert *ListNode to []int