Update solution 160 [README.md]

solution/160.Intersection of Two Linked Lists/README.md

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+## 相交链表
+### 题目描述
+
+编写一个程序，找到两个单链表相交的起始节点。
+
+
+示例 1：
+
+例如，下面的两个链表：
+
+A:          a1 → a2
+                   ↘
+                     c1 → c2 → c3
+                   ↗            
+B:     b1 → b2 → b3
+在节点 c1 开始相交。
+
+
+注意：
+
+如果两个链表没有交点，返回 null.
+在返回结果后，两个链表仍须保持原有的结构。
+可假定整个链表结构中没有循环。
+程序尽量满足 O(n) 时间复杂度，且仅用 O(1) 内存。
+
+### 解法
+第一遍循环，找出两个链表的长度差N
+第二遍循环，长链表先走N步，然后同时移动，判断是否有相同节点
+
+```java
+/**
+ * Definition for singly-linked list.
+ * public class ListNode {
+ *     int val;
+ *     ListNode next;
+ *     ListNode(int x) {
+ *         val = x;
+ *         next = null;
+ *     }
+ * }
+ */
+public class Solution {
+    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
+        if(headA == null || headB == null)
+            return null;
+        ListNode p = headA;
+        ListNode q = headB;
+        int pCount = 0;
+        int qCount = 0;
+        while(p.next != null || q.next != null){
+            if(p == q)
+                return p;
+
+            if(p.next != null)
+                p = p.next;
+            else
+                qCount++;
+
+            if(q.next != null)
+                q = q.next;
+            else
+                pCount++;
+        }
+        if(p != q)
+            return null;
+        p = headA;
+        q = headB;
+        while(pCount-- != 0){
+            p = p.next;
+        }
+        while(qCount-- != 0){
+            q = q.next;
+        }
+        while(p != q){
+            p = p.next;
+            q = q.next;
+        }
+
+        return p;
+
+
+    }
+}
+```

solution/160.Intersection of Two Linked Lists/Solution.java

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+/**
+ * Definition for singly-linked list.
+ * public class ListNode {
+ *     int val;
+ *     ListNode next;
+ *     ListNode(int x) {
+ *         val = x;
+ *         next = null;
+ *     }
+ * }
+ */
+public class Solution {
+    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
+        if(headA == null || headB == null)
+            return null;
+        ListNode p = headA;
+        ListNode q = headB;
+        int pCount = 0;
+        int qCount = 0;
+        while(p.next != null || q.next != null){
+            if(p == q)
+                return p;
+
+            if(p.next != null)
+                p = p.next;
+            else
+                qCount++;
+
+            if(q.next != null)
+                q = q.next;
+            else
+                pCount++;
+        }
+        if(p != q)
+            return null;
+        p = headA;
+        q = headB;
+        while(pCount-- != 0){
+            p = p.next;
+        }
+        while(qCount-- != 0){
+            q = q.next;
+        }
+        while(p != q){
+            p = p.next;
+            q = q.next;
+        }
+        return p;
+    }
+}