Merge pull request 0voice#54 from xiepeiyang/master

修复1.1.1答案错误

01.阿里篇/1.1.1 如何实现一个高效的单向链表逆序输出？.md

@@ -14,30 +14,26 @@ typedef struct node{
 
 void reverse(node* head)
 {
-    if(NULL == head || NULL == head->next){
+    if(head == NULL){
         return;
     }
-    
-    node* prev=NULL;
-    node* pcur=head->next;
-    node* next;
-    
-    while(pcur!=NULL){
-        if(pcur->next==NULL){
-            pcur->next=prev;
-            break;
-        }
-        next=pcur->next;
-        pcur->next=prev;
-        prev=pcur;
-        pcur=next;
+
+    node* pleft = NULL;
+    node* pcurrent = head;
+    node* pright = head->next;
+
+    while(pright){
+        pcurrent->next = pleft;
+        node *ptemp = pright->next;
+        pright->next = pcurrent;
+        pleft = pcurrent;
+        pcurrent = pright;
+        pright = ptemp;
     }
-    
-    head->next=pcur;
-    node*tmp=head->next;
-    while(tmp!=NULL){
-        cout<<tmp->data<<"\t";
-        tmp=tmp->next;
+
+    while(pcurrent != NULL){
+        cout<< pcurrent->data << "\t";
+        pcurrent = pcurrent->next;
     }
 }