编程导航算法村第二关

编程导航算法村第二关 | 白银挑战

指定区间的链表翻转

LeetCode92 ：给你单链表的头指针 head 和两个整数 left 和 right ，其中 left <= right。请你反转从位置 left 到位置 right 的链表节点，返回反转后的链表。

public ListNode reverseBetween(ListNode head, int left, int right) {
//        确定翻转列表的左起始节点ListNode domo = new ListNode(0);domo.next = head;
//        System.out.println(head);int i = 0;ListNode temp = domo;while (i < left - 1) {temp = temp.next;i++;}
//        System.out.println(temp.val);        保存前一个ListNode preNode = temp;ListNode leftNode = temp.next;        确定翻转列表的右结束节点i = 0;while (i < (right - left + 1)) {temp = temp.next;i++;}
//

        保存它的后一个ListNode rightNode = temp;ListNode afterNode = temp.next;preNode.next = null;rightNode.next = null;//        翻转指定区间列表temp = leftNode;ListNode pre = null;ListNode cul = leftNode;while (cul != null) {ListNode next = cul.next;cul.next = pre;pre = cul;cul = next;
//            temp = temp.next;}
//        System.out.println(rightNode.next.val);preNode.next = rightNode;leftNode.next = afterNode;return domo.next;}

两两交换链表

LeetCode24 给你一个链表，两两交换其中相邻的节点，并返回交换后链表的头节点。你必须在不修改节点内部的值的情况下完成本题

public ListNode swapPairs(ListNode head) {
//    设置虚拟头结点ListNode node = new ListNode(0);node.next = head;ListNode temp = node;
//        遍历到倒数第三个为止while (temp.next != null && temp.next.next != null) {ListNode n = temp.next.next;temp.next.next = n.next;n.next = temp.next;temp.next = n;temp = temp.next.next;}return node.next;}

实现单链表的+1

在此处，我们采取的思路是，现将链表进行翻转，然后给第一个节点+1，保存进位值，往后的节点与进位值相加
注意：我们需要考虑当遍历到最后一个元素还有进位的情况即999，这种情况，我们需要新建一个节点，插入到链表末尾
再次翻转元素

/*** 实现单链表所有元素+1的操作*/public static ListNode addOne(ListNode head) {if (head == null) {return null;}ListNode pre = reverse(head);ListNode temp = pre;int cut = 0; //进位
//        现将第一个元素添加了进位int sum = 0;sum = temp.val + 1;if (sum >= 10) {temp.val = sum % 10;cut = 1;} else {cut = 0;temp.val = sum;}temp = temp.next;while (temp != null) {sum = temp.val + cut;if (sum >= 10) {temp.val = sum % 10;cut = sum / 10;} else {temp.val = sum;cut = 0;}
//            到了最后一个，如果是最后一个还有进位，说明需要添加一位数if (temp.next == null) {if (cut == 1) {ListNode listNode = new ListNode(1);temp.next = listNode;break;}}temp = temp.next;}return reverse(pre);}/*** 定义翻转的函数*/public static ListNode reverse(ListNode head) {ListNode pre = null;ListNode cur = head;
//while (cur != null) {ListNode temp = cur.next;cur.next = pre;pre = cur;cur = temp;}return pre;}

链表加法

现将两个待相加的链表翻转，然后进行相加
分两种情况
- 位数相同
- 位数不同

 public ListNode addTwoNumbers(ListNode l1, ListNode l2) {if (l1 == null || l2 == null) {return null;}//         先翻转链表ListNode nl1 = reverse(l1);ListNode nl2 = reverse(l2);ListNode result = new ListNode(0);ListNode temp = result;int cul = 0;int sum = 0;
//                现将相同的位数进行相加while (nl1 != null && nl2 != null) {sum = nl1.val + nl2.val + cul;if (sum >= 10) {cul = 1;sum = sum % 10;} else {cul = 0;}ListNode node = new ListNode(sum);temp.next = node;temp = temp.next;nl1 = nl1.next;nl2 = nl2.next;//            考虑两者位数一样，同时到达末尾if (nl1 == null && nl2 == null) {if (cul != 0) {node = new ListNode(cul);temp.next = node;}}}//if (nl1 != null) {while (nl1 != null) {sum = nl1.val + cul;if (sum >= 10) {cul = sum / 10;sum = sum % 10;} else {cul = 0;}temp.next = new ListNode(sum);temp = temp.next;nl1 = nl1.next;if (nl1 == null && cul != 0) {temp.next = new ListNode(1);break;}}}if (nl2 != null) {while (nl2 != null) {sum = nl2.val + cul;if (sum >= 10) {cul = sum / 10;sum = sum % 10;} else {cul = 0;}temp.next = new ListNode(sum);temp = temp.next;nl2 = nl2.next;if (nl2 == null && cul != 0) {temp.next = new ListNode(1);break;}}}return reverse(result.next);}