finish 61

Author: BaffinLee

61. Rotate List.js

@@ -0,0 +1,68 @@
+/**
+ * 61. Rotate List
+ * 
+ * Given a linked list, rotate the list to the right by k places, where k is non-negative.
+ * 
+ * Example 1:
+ * 
+ * Input: 1->2->3->4->5->NULL, k = 2
+ * Output: 4->5->1->2->3->NULL
+ * Explanation:
+ * rotate 1 steps to the right: 5->1->2->3->4->NULL
+ * rotate 2 steps to the right: 4->5->1->2->3->NULL
+ * 
+ * Example 2:
+ * 
+ * Input: 0->1->2->NULL, k = 4
+ * Output: 2->0->1->NULL
+ * Explanation:
+ * rotate 1 steps to the right: 2->0->1->NULL
+ * rotate 2 steps to the right: 1->2->0->NULL
+ * rotate 3 steps to the right: 0->1->2->NULL
+ * rotate 4 steps to the right: 2->0->1->NULL
+ */
+
+/**
+ * Definition for singly-linked list.
+ * function ListNode(val) {
+ *     this.val = val;
+ *     this.next = null;
+ * }
+ */
+/**
+ * @param {ListNode} head
+ * @param {number} k
+ * @return {ListNode}
+ */
+var rotateRight = function(head, k) {
+	var count = 1;
+	var last = head;
+	var now = head;
+	
+	if (!head || !head.next) return head;
+
+	while (last.next) {
+		last = last.next;
+		count++;
+	}
+
+	k %= count;
+	
+	if (k === 0) return head;
+
+	while (k < count - 1) {
+		now = now.next;
+		k++;
+	}
+
+	last.next = head;
+	head = now.next;
+	now.next = null;
+
+	return head;
+};
+
+// 1. 拿到长度 count 和最后一个 last
+// 2. k %= count
+// 3. 找到新的最后一个 newLast，last.next = head，head = newLast.next，newLast.next = null
+// 4. return head