feat: add question 160

Author: yinxin630

160.相交链表.js

@@ -0,0 +1,95 @@
+/*
+ * @lc app=leetcode.cn id=160 lang=javascript
+ *
+ * [160] 相交链表
+ * 
+ * 有两种办法
+ * 1. 用 Set 存 A 中存在的节点, 然后遍历 B 查找是否有重合
+ * 2. 计算 A 和 B 的长度, 把较长的部分去掉使其等长, 然后 A 和 B 一起向下遍历
+ *    因为 A 和 B 是等长的, 那么如果有相交则一定是当前遍历到的 a 和 b 节点相等
+ *    如果找不到相等的节点, 那就是不想交
+ */
+
+// @lc code=start
+/**
+ * Definition for singly-linked list.
+ * function ListNode(val) {
+ *     this.val = val;
+ *     this.next = null;
+ * }
+ */
+
+/**
+ * @param {ListNode} headA
+ * @param {ListNode} headB
+ * @return {ListNode}
+ */
+/**
+ * Definition for singly-linked list.
+ * function ListNode(val) {
+ *     this.val = val;
+ *     this.next = null;
+ * }
+ */
+
+/**
+ * @param {ListNode} headA
+ * @param {ListNode} headB
+ * @return {ListNode}
+ */
+var getIntersectionNode = function(headA, headB) {
+    function getLen(head) {
+        let c = 0;
+        while (head) {
+            c++;
+            head = head.next;
+        }
+        return c;
+    }
+    const lenA = getLen(headA);
+    const lenB = getLen(headB);
+    if (lenA > lenB) {
+        let n = lenA - lenB;
+        while (n > 0) {
+            n--;
+            headA = headA.next;
+        }
+    } else {
+        let n = lenB - lenA;
+        while (n > 0) {
+            n--;
+            headB = headB.next;
+        }
+    }
+
+    let curA = headA, curB = headB;
+    while (curA && curB) {
+        if (curA === curB) {
+            return curA;
+        }
+        curA = curA.next;
+        curB = curB.next;
+    }
+    return null;
+};
+
+// var getIntersectionNode = function(headA, headB) {
+//     const set = new Set();
+//     let a = headA;
+//     while (a) {
+//         set.add(a);
+//         a = a.next;
+//     }
+
+//     let b = headB;
+//     while (b) {
+//         if (set.has(b)) {
+//             return b;
+//         }
+//         b = b.next;
+//     }
+
+//     return null;
+// };
+// @lc code=end
+