Create 143. Reorder List.md

solutions/143. Reorder List.md

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+# [143. Reorder List](https://leetcode.com/problems/reorder-list/)
+
+# 思路
+仔细分析题意可将整个解法分成三个部分:
+1. 找到链表中点位置(可用快慢指针实现), 并将链表从中点处断开，形成两个独立的链表;
+2. 翻转第二个链表;
+3. 将第二个链表的元素间隔地插入第一个链表中。
+
+## 改进
+上面的空间复杂度为O(1), 其实可以牺牲一下空间复杂度来加快速度: 用一个栈来存放后半部分节点, 这样就可以不用翻转链表而直接进行拼接了.
+
+# C++
+```
+class Solution {
+public:
+    void reorderList(ListNode* head) {
+        ListNode *fast = head, *slow = head;
+        while(fast && fast -> next){
+            fast = fast -> next -> next;
+            slow = slow -> next;
+        }
+        if(slow == fast) return; // 不到三个节点
+        
+        // 翻转链表的后半部分
+        ListNode *tmp1 = slow -> next, *tmp2;
+        slow -> next = NULL;
+        while(tmp1){
+            tmp2 = tmp1 -> next;
+            tmp1 -> next = slow -> next;
+            slow -> next = tmp1;
+            tmp1 = tmp2;
+        }
+
+        ListNode *p1 = head, *p2 = slow -> next;
+        slow -> next = NULL;
+        while(p2){
+            tmp1 = p1 -> next;
+            tmp2 = p2 -> next;
+            p1 -> next = p2;
+            p2 -> next = tmp1;
+            
+            p1 = p2 -> next;
+            p2 = tmp2;
+        }
+    }
+};
+```
+
+## 改进
+```C++
+class Solution {
+public:
+    void reorderList(ListNode* head) {
+        ListNode *fast = head, *slow = head;
+        while(fast && fast -> next){
+            fast = fast -> next -> next;
+            slow = slow -> next;
+        }
+        if(slow == fast) return; // 不到三个节点
+        
+        // 用一个栈存放后半部分
+        stack<ListNode *>stk;
+        ListNode *tmp1 = slow -> next;
+        slow -> next = NULL;
+        while(tmp1){
+            stk.push(tmp1);
+            tmp1 = tmp1 -> next;
+        }
+
+        ListNode *p1 = head, *p2;
+        while(!stk.empty()){
+            p2 = stk.top();
+            stk.pop();
+            
+            tmp1 = p1 -> next;
+            p1 -> next = p2;
+            p2 -> next = tmp1;
+            
+            p1 = tmp1;
+        }
+    }
+};
+```