修正 LinkList.remove 方法没有更新 prevnode 的 bug

Author: PegasusWang

docs/03_链表/linked_list.py

@@ -74,6 +74,8 @@ def remove(self, value):    # O(n)
                 del curnode
                 self.length -= 1
                 return 1  # 表明删除成功
+            else:
+                prevnode = curnode
         return -1  # 表明删除失败
 
     def find(self, value):    # O(n)
@@ -113,26 +115,27 @@ def test_linked_list():
     ll.append(0)
     ll.append(1)
     ll.append(2)
+    ll.append(3)
 
-    assert len(ll) == 3
+    assert len(ll) == 4
     assert ll.find(2) == 2
-    assert ll.find(3) == -1
+    assert ll.find(-1) == -1
 
     assert ll.remove(0) == 1
-    assert ll.remove(3) == -1
+    assert ll.remove(10) == -1
+    assert ll.remove(2) == 1
     assert len(ll) == 2
+    assert list(ll) == [1, 3]
     assert ll.find(0) == -1
 
-    assert list(ll) == [1, 2]
-
     ll.appendleft(0)
-    assert list(ll) == [0, 1, 2]
+    assert list(ll) == [0, 1, 3]
     assert len(ll) == 3
 
     headvalue = ll.popleft()
     assert headvalue == 0
     assert len(ll) == 2
-    assert list(ll) == [1, 2]
+    assert list(ll) == [1, 3]
 
     ll.clear()
     assert len(ll) == 0

docs/04_队列/queue.py

@@ -69,12 +69,15 @@ def remove(self, value):    # O(n)
         """
         prevnode = self.root    #
         curnode = self.root.next
-        while curnode.next is not None:
+        for curnode in self.iter_node():
             if curnode.value == value:
                 prevnode.next = curnode.next
                 del curnode
                 self.length -= 1
-                return
+                return 1  # 表明删除成功
+            else:
+                prevnode = curnode
+        return -1  # 表明删除失败
 
     def find(self, value):    # O(n)
         """ 查找一个节点，返回序号，从 0 开始