From c39b7a328f4284d9bc96740baeb221c4f0a488ca Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E5=94=90=E6=A0=91=E6=A3=AE?= <tangshusen@pku.edu.cn>
Date: Sat, 2 Feb 2019 21:52:01 +0800
Subject: [PATCH] Create 61. Rotate List.md

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+# [61. Rotate List](https://leetcode.com/problems/rotate-list/)
+# 思路
+题目要求循环移位链表k步。    
+不要被题目迷惑了，这题根本不需要移位，只需要在倒数第k个位置处将链表截断然后将后半部分接到最前面就可以了。现在问题就是怎样快速找到倒数第k个的位置。     
+涉及到链表倒数某个位置的一般就是用一个双指针pre和p，初始都为head，先让p后移k步再让pre和p同时后移，这样当p到达链尾时pre就处于倒数第k个位置。    
+注意k可能大于链表的长度，所以要先将k对链表长度len取模。         
+时间复杂度O(n)，空间复杂度O(1)
+
+# C++
+``` C++
+/**
+ * Definition for singly-linked list.
+ * struct ListNode {
+ *     int val;
+ *     ListNode *next;
+ *     ListNode(int x) : val(x), next(NULL) {}
+ * };
+ */
+class Solution {
+public:
+    ListNode* rotateRight(ListNode* head, int k) {
+        if(!head) return head;
+        int len = 0;
+        ListNode *p = head, *pre=head, *res;
+        while(p){
+            len++;
+            p = p -> next;
+        } // 获得链表长度
+        k %= len;
+        if(k == 0) return head; 
+        p = head;
+        while(k--) p = p -> next; // p先后移k步
+        while(p -> next){ // pre和p同时后移直到p -> next为NULL
+            p = p -> next;
+            pre = pre -> next;
+        } // 此时pre -> next 就是倒数第k个元素
+        res = pre -> next;
+        pre -> next = NULL;
+        p -> next = head;
+        return res;
+    }
+};
+```