402 lines
12 KiB
Java
402 lines
12 KiB
Java
<!-- GFM-TOC -->
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* [20. 表示数值的字符串](#20-表示数值的字符串)
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* [21. 调整数组顺序使奇数位于偶数前面](#21-调整数组顺序使奇数位于偶数前面)
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* [22. 链表中倒数第 K 个结点](#22-链表中倒数第-k-个结点)
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* [23. 链表中环的入口结点](#23-链表中环的入口结点)
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* [24. 反转链表](#24-反转链表)
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* [25. 合并两个排序的链表](#25-合并两个排序的链表)
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* [26. 树的子结构](#26-树的子结构)
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* [27. 二叉树的镜像](#27-二叉树的镜像)
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* [28 对称的二叉树](#28-对称的二叉树)
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* [29. 顺时针打印矩阵](#29-顺时针打印矩阵)
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<!-- GFM-TOC -->
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# 20. 表示数值的字符串
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[NowCoder](https://www.nowcoder.com/practice/6f8c901d091949a5837e24bb82a731f2?tpId=13&tqId=11206&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking)
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## 题目描述
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```
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true
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"+100"
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"5e2"
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"-123"
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"3.1416"
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"-1E-16"
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```
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```
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false
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"12e"
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"1a3.14"
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"1.2.3"
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"+-5"
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"12e+4.3"
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```
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## 解题思路
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使用正则表达式进行匹配。
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```html
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[] ： 字符集合
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() ： 分组
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? ： 重复 0 ~ 1 次
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+ ： 重复 1 ~ n 次
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* ： 重复 0 ~ n 次
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. ： 任意字符
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\\. ： 转义后的 .
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\\d ： 数字
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```
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```java
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public boolean isNumeric(char[] str) {
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if (str == null || str.length == 0)
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return false;
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return new String(str).matches("[+-]?\\d*(\\.\\d+)?([eE][+-]?\\d+)?");
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}
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```
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# 21. 调整数组顺序使奇数位于偶数前面
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[NowCoder](https://www.nowcoder.com/practice/beb5aa231adc45b2a5dcc5b62c93f593?tpId=13&tqId=11166&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking)
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## 题目描述
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需要保证奇数和奇数，偶数和偶数之间的相对位置不变，这和书本不太一样。
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<div align="center"> <img src="pics/d03a2efa-ef19-4c96-97e8-ff61df8061d3.png" width="200px"> </div><br>
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## 解题思路
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方法一：创建一个新数组，时间复杂度 O(N)，空间复杂度 O(N)。
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```java
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public void reOrderArray(int[] nums) {
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// 奇数个数
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int oddCnt = 0;
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for (int x : nums)
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if (!isEven(x))
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oddCnt++;
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int[] copy = nums.clone();
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int i = 0, j = oddCnt;
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for (int num : copy) {
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if (num % 2 == 1)
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nums[i++] = num;
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else
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nums[j++] = num;
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}
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}
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private boolean isEven(int x) {
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return x % 2 == 0;
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}
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```
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方法二：使用冒泡思想，每次都当前偶数上浮到当前最右边。时间复杂度 O(N<sup>2</sup>)，空间复杂度 O(1)，时间换空间。
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```java
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public void reOrderArray(int[] nums) {
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int N = nums.length;
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for (int i = N - 1; i > 0; i--) {
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for (int j = 0; j < i; j++) {
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if (isEven(nums[j]) && !isEven(nums[j + 1])) {
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swap(nums, j, j + 1);
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}
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}
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}
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}
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private boolean isEven(int x) {
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return x % 2 == 0;
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}
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private void swap(int[] nums, int i, int j) {
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int t = nums[i];
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nums[i] = nums[j];
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nums[j] = t;
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}
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```
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# 22. 链表中倒数第 K 个结点
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[NowCoder](https://www.nowcoder.com/practice/529d3ae5a407492994ad2a246518148a?tpId=13&tqId=11167&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking)
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## 解题思路
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设链表的长度为 N。设置两个指针 P1 和 P2，先让 P1 移动 K 个节点，则还有 N - K 个节点可以移动。此时让 P1 和 P2 同时移动，可以知道当 P1 移动到链表结尾时，P2 移动到第 N - K 个节点处，该位置就是倒数第 K 个节点。
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<div align="center"> <img src="pics/6b504f1f-bf76-4aab-a146-a9c7a58c2029.png" width="500"/> </div><br>
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```java
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public ListNode FindKthToTail(ListNode head, int k) {
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if (head == null)
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return null;
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ListNode P1 = head;
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while (P1 != null && k-- > 0)
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P1 = P1.next;
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if (k > 0)
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return null;
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ListNode P2 = head;
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while (P1 != null) {
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P1 = P1.next;
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P2 = P2.next;
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}
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return P2;
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}
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```
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# 23. 链表中环的入口结点
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[NowCoder](https://www.nowcoder.com/practice/253d2c59ec3e4bc68da16833f79a38e4?tpId=13&tqId=11208&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking)
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## 题目描述
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一个链表中包含环，请找出该链表的环的入口结点。要求不能使用额外的空间。
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## 解题思路
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使用双指针，一个指针 fast 每次移动两个节点，一个指针 slow 每次移动一个节点。因为存在环，所以两个指针必定相遇在环中的某个节点上。假设相遇点在下图的 z1 位置，此时 fast 移动的节点数为 x+2y+z，slow 为 x+y，由于 fast 速度比 slow 快一倍，因此 x+2y+z=2(x+y)，得到 x=z。
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在相遇点，slow 要到环的入口点还需要移动 z 个节点，如果让 fast 重新从头开始移动，并且速度变为每次移动一个节点，那么它到环入口点还需要移动 x 个节点。在上面已经推导出 x=z，因此 fast 和 slow 将在环入口点相遇。
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<div align="center"> <img src="pics/bb7fc182-98c2-4860-8ea3-630e27a5f29f.png" width="500"/> </div><br>
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```java
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public ListNode EntryNodeOfLoop(ListNode pHead) {
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if (pHead == null || pHead.next == null)
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return null;
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ListNode slow = pHead, fast = pHead;
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do {
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fast = fast.next.next;
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slow = slow.next;
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} while (slow != fast);
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fast = pHead;
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while (slow != fast) {
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slow = slow.next;
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fast = fast.next;
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}
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return slow;
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}
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```
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# 24. 反转链表
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[NowCoder](https://www.nowcoder.com/practice/75e878df47f24fdc9dc3e400ec6058ca?tpId=13&tqId=11168&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking)
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## 解题思路
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### 递归
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```java
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public ListNode ReverseList(ListNode head) {
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if (head == null || head.next == null)
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return head;
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ListNode next = head.next;
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head.next = null;
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ListNode newHead = ReverseList(next);
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next.next = head;
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return newHead;
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}
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```
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### 迭代
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使用头插法。
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```java
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public ListNode ReverseList(ListNode head) {
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ListNode newList = new ListNode(-1);
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while (head != null) {
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ListNode next = head.next;
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head.next = newList.next;
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newList.next = head;
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head = next;
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}
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return newList.next;
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}
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```
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# 25. 合并两个排序的链表
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[NowCoder](https://www.nowcoder.com/practice/d8b6b4358f774294a89de2a6ac4d9337?tpId=13&tqId=11169&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking)
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## 题目描述
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<div align="center"> <img src="pics/c094d2bc-ec75-444b-af77-d369dfb6b3b4.png" width="400"/> </div><br>
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## 解题思路
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### 递归
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```java
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public ListNode Merge(ListNode list1, ListNode list2) {
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if (list1 == null)
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return list2;
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if (list2 == null)
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return list1;
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if (list1.val <= list2.val) {
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list1.next = Merge(list1.next, list2);
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return list1;
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} else {
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list2.next = Merge(list1, list2.next);
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return list2;
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}
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}
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```
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### 迭代
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```java
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public ListNode Merge(ListNode list1, ListNode list2) {
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ListNode head = new ListNode(-1);
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ListNode cur = head;
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while (list1 != null && list2 != null) {
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if (list1.val <= list2.val) {
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cur.next = list1;
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list1 = list1.next;
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} else {
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cur.next = list2;
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list2 = list2.next;
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}
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cur = cur.next;
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}
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if (list1 != null)
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cur.next = list1;
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if (list2 != null)
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cur.next = list2;
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return head.next;
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}
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```
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# 26. 树的子结构
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[NowCoder](https://www.nowcoder.com/practice/6e196c44c7004d15b1610b9afca8bd88?tpId=13&tqId=11170&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking)
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## 题目描述
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<div align="center"> <img src="pics/84a5b15a-86c5-4d8e-9439-d9fd5a4699a1.jpg" width="450"/> </div><br>
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## 解题思路
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```java
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public boolean HasSubtree(TreeNode root1, TreeNode root2) {
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if (root1 == null || root2 == null)
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return false;
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return isSubtreeWithRoot(root1, root2) || HasSubtree(root1.left, root2) || HasSubtree(root1.right, root2);
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}
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private boolean isSubtreeWithRoot(TreeNode root1, TreeNode root2) {
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if (root2 == null)
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return true;
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if (root1 == null)
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return false;
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if (root1.val != root2.val)
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return false;
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return isSubtreeWithRoot(root1.left, root2.left) && isSubtreeWithRoot(root1.right, root2.right);
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}
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```
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# 27. 二叉树的镜像
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[NowCoder](https://www.nowcoder.com/practice/564f4c26aa584921bc75623e48ca3011?tpId=13&tqId=11171&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking)
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## 题目描述
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<div align="center"> <img src="pics/0c12221f-729e-4c22-b0ba-0dfc909f8adf.jpg" width="300"/> </div><br>
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## 解题思路
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```java
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public void Mirror(TreeNode root) {
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if (root == null)
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return;
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swap(root);
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Mirror(root.left);
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Mirror(root.right);
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}
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private void swap(TreeNode root) {
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TreeNode t = root.left;
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root.left = root.right;
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root.right = t;
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}
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```
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# 28 对称的二叉树
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[NowCoder](https://www.nowcoder.com/practice/ff05d44dfdb04e1d83bdbdab320efbcb?tpId=13&tqId=11211&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking)
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## 题目描述
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<div align="center"> <img src="pics/0c12221f-729e-4c22-b0ba-0dfc909f8adf.jpg" width="300"/> </div><br>
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## 解题思路
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```java
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boolean isSymmetrical(TreeNode pRoot) {
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if (pRoot == null)
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return true;
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return isSymmetrical(pRoot.left, pRoot.right);
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}
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boolean isSymmetrical(TreeNode t1, TreeNode t2) {
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if (t1 == null && t2 == null)
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return true;
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if (t1 == null || t2 == null)
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return false;
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if (t1.val != t2.val)
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return false;
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return isSymmetrical(t1.left, t2.right) && isSymmetrical(t1.right, t2.left);
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}
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```
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# 29. 顺时针打印矩阵
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[NowCoder](https://www.nowcoder.com/practice/9b4c81a02cd34f76be2659fa0d54342a?tpId=13&tqId=11172&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking)
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## 题目描述
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下图的矩阵顺时针打印结果为：1, 2, 3, 4, 8, 12, 16, 15, 14, 13, 9, 5, 6, 7, 11, 10
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<div align="center"> <img src="pics/48517227-324c-4664-bd26-a2d2cffe2bfe.png" width="200px"> </div><br>
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## 解题思路
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```java
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public ArrayList<Integer> printMatrix(int[][] matrix) {
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ArrayList<Integer> ret = new ArrayList<>();
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int r1 = 0, r2 = matrix.length - 1, c1 = 0, c2 = matrix[0].length - 1;
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while (r1 <= r2 && c1 <= c2) {
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for (int i = c1; i <= c2; i++)
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ret.add(matrix[r1][i]);
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for (int i = r1 + 1; i <= r2; i++)
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ret.add(matrix[i][c2]);
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if (r1 != r2)
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for (int i = c2 - 1; i >= c1; i--)
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ret.add(matrix[r2][i]);
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if (c1 != c2)
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for (int i = r2 - 1; i > r1; i--)
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ret.add(matrix[i][c1]);
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r1++; r2--; c1++; c2--;
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}
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return ret;
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}
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```
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# 微信公众号
|
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|
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|
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更多精彩内容将发布在微信公众号 CyC2018 上，你也可以在公众号后台和我交流学习和求职相关的问题。另外，公众号提供了该项目的 PDF 等离线阅读版本，后台回复 "下载" 即可领取。公众号也提供了一份技术面试复习大纲，不仅系统整理了面试知识点，而且标注了各个知识点的重要程度，从而帮你理清多而杂的面试知识点，后台回复 "大纲" 即可领取。我基本是按照这个大纲来进行复习的，对我拿到了 BAT 头条等 Offer 起到很大的帮助。你们完全可以和我一样根据大纲上列的知识点来进行复习，就不用看很多不重要的内容，也可以知道哪些内容很重要从而多安排一些复习时间。
|
||
|
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|
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<br><div align="center"><img width="320px" src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/other/公众号海报6.png"></img></div>
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