leetcode 92 reverse between

src/main/java/list/LeetCode_0092_ReverseLinkedListII.java

-
-package list;
-
-// 反转单链表一部分
-// https://leetcode.cn/problems/reverse-linked-list-ii/
-// 笔记：https://www.cnblogs.com/greyzeng/p/16629407.html
-public class LeetCode_0092_ReverseLinkedListII {
-    public static class ListNode {
-        int val;
-        ListNode next;
-
-        ListNode() {
-        }
-
-        ListNode(int val) {
-            this.val = val;
-        }
-
-        ListNode(int val, ListNode next) {
-            this.val = val;
-            this.next = next;
-        }
-    }
-
-    public static ListNode reverseBetween(ListNode head, int m, int n) {
-        if (m == n) {
-            // 不变
-            return head;
-        }
-        ListNode startPre = null;
-        ListNode start = null;
-        ListNode end;
-        ListNode endAfter = null;
-        ListNode cur = head;
-        int i = 1;
-        while (i <= n) {
-            if (i == m - 1) {
-                startPre = cur;
-            } else if (i == m) {
-                start = cur;
-            } else if (i == n) {
-                end = cur;
-                if (end.next != null) {
-                    endAfter = end.next;
-                    end.next = null;
-                }
-            }
-            cur = cur.next;
-            i++;
-        }
-        i = m;
-        ListNode pre = null;
-        // 反转链表操作
-        while (i <= n) {
-            ListNode tmp = start.next;
-            start.next = pre;
-            pre = start;
-            start = tmp;
-            if (i == m) {
-                pre.next = endAfter;
-            }
-            i++;
-        }
-        // 换头了
-        if (m == 1 && n != 1) {
-            return pre;
-        }
-        startPre.next = pre;
-        return head;
-    }
-
-    public static ListNode reverseBetweenRecursive(ListNode head, int m, int n) {
-        if (m == 1) {
-            return reverseN(head, n);
-        }
-        head.next = reverseBetweenRecursive(head.next, m - 1, n - 1);
-        return head;
-    }
-
-    static ListNode successor = null;
-
-    // 反转链表前N个节点
-    public static ListNode reverseN(ListNode head, int n) {
-        if (n == 1) {
-            successor = head.next;
-            return head;
-        }
-        ListNode last = reverseN(head.next, n - 1);
-        head.next.next = head;
-        head.next = successor;
-        return last;
-    }
-
-    public static void main(String[] args) {
-        testCase1();
-        testCase2();
-    }
-
-    public static void testCase1() {
-        System.out.println("begin test");
-        ListNode head = new ListNode(1);
-        head.next = new ListNode(2);
-        head.next.next = new ListNode(3);
-        head.next.next.next = new ListNode(4);
-        head.next.next.next.next = new ListNode(5);
-        head.next.next.next.next.next = new ListNode(6);
-        ListNode t = reverseBetween(head, 2, 4);
-        while (t != null) {
-            System.out.print(t.val + " ");
-            t = t.next;
-        }
-        System.out.println("end test");
-    }
-
-    public static void testCase2() {
-        System.out.println("begin test");
-        ListNode head = new ListNode(3);
-        head.next = new ListNode(5);
-
-        ListNode t = reverseBetween(head, 1, 2);
-        while (t != null) {
-            System.out.print(t.val + " ");
-            t = t.next;
-        }
-        System.out.println("end test");
-    }
-}

src/main/java/resolved/list/LeetCode_0092_ReverseLinkedListII.java

+package resolved.list;
+
+// 值得反复练习的习题
+// 反转单链表一部分
+// https://leetcode.cn/problems/reverse-linked-list-ii/
+// 笔记：https://www.cnblogs.com/greyzeng/p/16629407.html
+public class LeetCode_0092_ReverseLinkedListII {
+  public static class ListNode {
+    int val;
+    ListNode next;
+
+    ListNode() {}
+
+    ListNode(int val) {
+      this.val = val;
+    }
+
+    ListNode(int val, ListNode next) {
+      this.val = val;
+      this.next = next;
+    }
+  }
+
+  public ListNode reverseBetween(ListNode h, int m, int n) {
+    if (m == n || h == null || h.next == null) {
+      return h;
+    }
+    ListNode dummy = new ListNode(0, h);
+    ListNode startPre = dummy;
+    ListNode start = h;
+    for (int i = 1; i < m; i++) {
+      startPre = startPre.next;
+      start = start.next;
+    }
+    ListNode pre = null;
+    ListNode cur = start;
+    for (int i = m; i <= n; i++) {
+      ListNode tmp = cur.next;
+      cur.next = pre;
+      pre = cur;
+      cur = tmp;
+    }
+    if (m != 1) {
+      startPre.next = pre;
+      start.next = cur;
+      return h;
+    } else {
+      start.next = cur;
+      return pre;
+    }
+  }
+
+  public static ListNode reverseBetweenRecursive(ListNode head, int m, int n) {
+    if (m == 1) {
+      return reverseN(head, n);
+    }
+    head.next = reverseBetweenRecursive(head.next, m - 1, n - 1);
+    return head;
+  }
+
+  static ListNode successor = null;
+
+  // 反转链表前N个节点
+  public static ListNode reverseN(ListNode head, int n) {
+    if (n == 1) {
+      successor = head.next;
+      return head;
+    }
+    ListNode last = reverseN(head.next, n - 1);
+    head.next.next = head;
+    head.next = successor;
+    return last;
+  }
+}