diff --git a/160. Intersection of Two Linked Lists.md b/160. Intersection of Two Linked Lists.md
new file mode 100644
index 0000000..5743db3
--- /dev/null
+++ b/160. Intersection of Two Linked Lists.md	
@@ -0,0 +1,52 @@
+# [160. Intersection of Two Linked Lists](https://leetcode.com/problems/intersection-of-two-linked-lists/description/)
+# 思路
+题目要求求两个链表的交集，需要注意的事实是只要两个链表有一个节点重合了，那么其后的节点也是重合的，这个节点也即所求，就如题图c1所示。   
+先计算两个链表的长度差delta_len，然后设置两个工作指针p1和p2，保证p1所在的链表是较短的链表, 再让p2先往后移动delta_len步，最后p1、p2同时往后移动
+直到p1 == p2即到达所求节点c1或者遇到NULL。   
+时间复杂度O(n), 空间复杂度O(1)
+# C++
+```
+/**
+ * Definition for singly-linked list.
+ * struct ListNode {
+ *     int val;
+ *     ListNode *next;
+ *     ListNode(int x) : val(x), next(NULL) {}
+ * };
+ */
+class Solution {
+public:
+    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
+        if(!(headA && headB)) return NULL;
+        int delta_len = 0;
+        ListNode *p1 = headA, *p2 = headB, *tmp = NULL;
+        while(p1 && p2){
+            p1 = p1 -> next;
+            p2 = p2 -> next;
+        }
+        if(p1) tmp = p1;
+        else tmp = p2;
+        while(tmp){
+            delta_len++;
+            tmp = tmp -> next;
+        }
+        //以上代码求链表长度差
+          
+        if(p1){ // 保证p1长度不大于p2
+            p1 = headB;
+            p2 = headA;
+        }
+        else{
+            p1 = headA;
+            p2 = headB;
+        }
+        
+        while(delta_len--) p2 = p2 -> next;
+        while(p1 && p1 != p2){
+            p1 = p1 -> next;
+            p2 = p2 -> next;
+        }
+        return p1;
+    }
+};
+```