From c1899dd27071ab3573c659c64a44fbae8e7925fa Mon Sep 17 00:00:00 2001
From: zhangsan <646228430@qq.com>
Date: Fri, 20 Jun 2025 13:28:35 +0800
Subject: [PATCH] =?UTF-8?q?6.20=20=E4=BA=8C=E5=88=B7hot100=20=E7=9F=A9?=
 =?UTF-8?q?=E9=98=B5+=E9=93=BE=E8=A1=A8=E5=BC=80=E5=A4=B4?=
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---
 .../java/linkedlist/GetIntersectionNode.java  | 44 +++++++++++++++++--
 src/main/java/linkedlist/ReverseList.java     |  1 +
 src/main/java/matrix/RotateImage.java         |  2 +
 src/main/java/matrix/SearchMatrix.java        |  1 +
 4 files changed, 44 insertions(+), 4 deletions(-)

diff --git a/src/main/java/linkedlist/GetIntersectionNode.java b/src/main/java/linkedlist/GetIntersectionNode.java
index 271d489..b00f969 100644
--- a/src/main/java/linkedlist/GetIntersectionNode.java
+++ b/src/main/java/linkedlist/GetIntersectionNode.java
@@ -5,14 +5,16 @@ import java.util.HashSet;
 /**
  * 题目：160. 相交链表  (getIntersectionNode)
  * 描述：给你两个单链表的头节点 headA 和 headB ，请你找出并返回两个单链表相交的起始节点。如果两个链表不存在相交节点，返回 null 。
-
+ *
+ 进阶：你能否设计一个时间复杂度 O(m + n) 、仅用 O(1) 内存的解决方案？
  * 链接：https://leetcode.cn/problems/intersection-of-two-linked-lists/
 
  */
 
 //没想出高效方法
+//二刷会做
 public class GetIntersectionNode {
-    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
+    public ListNode getIntersectionNode1(ListNode headA, ListNode headB) {
         HashSet<ListNode> visited = new HashSet<ListNode>();
         ListNode temp = headA;
         while (temp != null) {
@@ -28,8 +30,42 @@ public class GetIntersectionNode {
         }
         return null;
     }
-    //高效方法
-    public ListNode getIntersectionNode1(ListNode headA, ListNode headB) {
+    //二刷想到的，时间复杂度O(m + n)，但是有点啰嗦
+    public ListNode getIntersectionNode2(ListNode headA, ListNode headB) {
+        int lenA = 0, lenB = 0;
+        ListNode curA = headA, curB = headB;
+
+        // 1. 统计长度
+        while (curA != null) {
+            lenA++;
+            curA = curA.next;
+        }
+        while (curB != null) {
+            lenB++;
+            curB = curB.next;
+        }
+
+        // 2. 让长的链表先走差值步
+        curA = headA;
+        curB = headB;
+        if (lenA > lenB) {
+            for (int i = 0; i < lenA - lenB; i++) curA = curA.next;
+        } else {
+            for (int i = 0; i < lenB - lenA; i++) curB = curB.next;
+        }
+
+        // 3. 同时前进，找相遇点
+        while (curA != null && curB != null) {
+            if (curA == curB) return curA; // 找到相交点
+            curA = curA.next;
+            curB = curB.next;
+        }
+
+        return null; // 不相交
+    }
+
+    //高效方法，链表 A 的长度为 m = a + c（a 是独有部分，c 是公共部分）  链表 B 的长度为 n = b + c  要么相遇在交点，要么都走到 null。
+    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
         ListNode A = headA, B = headB;
         while (A != B) {
             A = A != null ? A.next : headB;
diff --git a/src/main/java/linkedlist/ReverseList.java b/src/main/java/linkedlist/ReverseList.java
index 9549180..cd92b67 100644
--- a/src/main/java/linkedlist/ReverseList.java
+++ b/src/main/java/linkedlist/ReverseList.java
@@ -6,6 +6,7 @@ package linkedlist;
  * 链接：https://leetcode.cn/problems/reverse-linked-list/
 
  */
+//二刷会做
 public class ReverseList {
     public ListNode reverseList(ListNode head) {
         ListNode prev = null;
diff --git a/src/main/java/matrix/RotateImage.java b/src/main/java/matrix/RotateImage.java
index 0e289a1..cfd93b2 100644
--- a/src/main/java/matrix/RotateImage.java
+++ b/src/main/java/matrix/RotateImage.java
@@ -15,7 +15,9 @@ package matrix;
  输出：[[15,13,2,5],[14,3,4,1],[12,6,8,9],[16,7,10,11]]
  */
 //不会原地旋转，只会借助辅助数组
+//二刷不会
 public class RotateImage {
+    //[i][j] → [j][n - 1 - i]
     public void rotate(int[][] matrix) {
         int n = matrix.length;
         int[][] matrix_new = new int[n][n];
diff --git a/src/main/java/matrix/SearchMatrix.java b/src/main/java/matrix/SearchMatrix.java
index fc2508f..3e1da42 100644
--- a/src/main/java/matrix/SearchMatrix.java
+++ b/src/main/java/matrix/SearchMatrix.java
@@ -16,6 +16,7 @@ package matrix;
  输出：false
  */
 //高效的方法不会
+//二刷会做
 public class SearchMatrix {
     //矩阵逆时针旋转45，右上角的元素看作树根，左边比他小，右边比他大
     public boolean searchMatrix(int[][] matrix, int target) {