397 lines
12 KiB
Markdown
397 lines
12 KiB
Markdown
<!-- GFM-TOC -->
|
||
* [20. 表示数值的字符串](#20-表示数值的字符串)
|
||
* [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)
|
||
* [22. 链表中倒数第 K 个结点](#22-链表中倒数第-k-个结点)
|
||
* [23. 链表中环的入口结点](#23-链表中环的入口结点)
|
||
* [24. 反转链表](#24-反转链表)
|
||
* [25. 合并两个排序的链表](#25-合并两个排序的链表)
|
||
* [26. 树的子结构](#26-树的子结构)
|
||
* [27. 二叉树的镜像](#27-二叉树的镜像)
|
||
* [28 对称的二叉树](#28-对称的二叉树)
|
||
* [29. 顺时针打印矩阵](#29-顺时针打印矩阵)
|
||
<!-- GFM-TOC -->
|
||
|
||
|
||
# 20. 表示数值的字符串
|
||
|
||
[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)
|
||
|
||
## 题目描述
|
||
|
||
```
|
||
true
|
||
|
||
"+100"
|
||
"5e2"
|
||
"-123"
|
||
"3.1416"
|
||
"-1E-16"
|
||
```
|
||
|
||
```
|
||
false
|
||
|
||
"12e"
|
||
"1a3.14"
|
||
"1.2.3"
|
||
"+-5"
|
||
"12e+4.3"
|
||
```
|
||
|
||
|
||
## 解题思路
|
||
|
||
使用正则表达式进行匹配。
|
||
|
||
```html
|
||
[] ： 字符集合
|
||
() ： 分组
|
||
? ： 重复 0 ~ 1 次
|
||
+ ： 重复 1 ~ n 次
|
||
* ： 重复 0 ~ n 次
|
||
. ： 任意字符
|
||
\\. ： 转义后的 .
|
||
\\d ： 数字
|
||
```
|
||
|
||
```java
|
||
public boolean isNumeric(char[] str) {
|
||
if (str == null || str.length == 0)
|
||
return false;
|
||
return new String(str).matches("[+-]?\\d*(\\.\\d+)?([eE][+-]?\\d+)?");
|
||
}
|
||
```
|
||
|
||
# 21. 调整数组顺序使奇数位于偶数前面
|
||
|
||
[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)
|
||
|
||
## 题目描述
|
||
|
||
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。
|
||
|
||
<div align="center"> <img src="pics/d03a2efa-ef19-4c96-97e8-ff61df8061d3.png" width="200px"> </div><br>
|
||
|
||
## 解题思路
|
||
|
||
方法一：创建一个新数组，时间复杂度 O(N)，空间复杂度 O(N)。
|
||
|
||
```java
|
||
public void reOrderArray(int[] nums) {
|
||
// 奇数个数
|
||
int oddCnt = 0;
|
||
for (int x : nums)
|
||
if (!isEven(x))
|
||
oddCnt++;
|
||
int[] copy = nums.clone();
|
||
int i = 0, j = oddCnt;
|
||
for (int num : copy) {
|
||
if (num % 2 == 1)
|
||
nums[i++] = num;
|
||
else
|
||
nums[j++] = num;
|
||
}
|
||
}
|
||
|
||
private boolean isEven(int x) {
|
||
return x % 2 == 0;
|
||
}
|
||
```
|
||
|
||
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。
|
||
|
||
```java
|
||
public void reOrderArray(int[] nums) {
|
||
int N = nums.length;
|
||
for (int i = N - 1; i > 0; i--) {
|
||
for (int j = 0; j < i; j++) {
|
||
if (isEven(nums[j]) && !isEven(nums[j + 1])) {
|
||
swap(nums, j, j + 1);
|
||
}
|
||
}
|
||
}
|
||
}
|
||
|
||
private boolean isEven(int x) {
|
||
return x % 2 == 0;
|
||
}
|
||
|
||
private void swap(int[] nums, int i, int j) {
|
||
int t = nums[i];
|
||
nums[i] = nums[j];
|
||
nums[j] = t;
|
||
}
|
||
```
|
||
|
||
# 22. 链表中倒数第 K 个结点
|
||
|
||
[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)
|
||
|
||
## 解题思路
|
||
|
||
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。
|
||
|
||
<div align="center"> <img src="pics/6b504f1f-bf76-4aab-a146-a9c7a58c2029.png" width="500"/> </div><br>
|
||
|
||
```java
|
||
public ListNode FindKthToTail(ListNode head, int k) {
|
||
if (head == null)
|
||
return null;
|
||
ListNode P1 = head;
|
||
while (P1 != null && k-- > 0)
|
||
P1 = P1.next;
|
||
if (k > 0)
|
||
return null;
|
||
ListNode P2 = head;
|
||
while (P1 != null) {
|
||
P1 = P1.next;
|
||
P2 = P2.next;
|
||
}
|
||
return P2;
|
||
}
|
||
```
|
||
|
||
# 23. 链表中环的入口结点
|
||
|
||
[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)
|
||
|
||
## 题目描述
|
||
|
||
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。
|
||
|
||
## 解题思路
|
||
|
||
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。
|
||
|
||
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。
|
||
|
||
<div align="center"> <img src="pics/bb7fc182-98c2-4860-8ea3-630e27a5f29f.png" width="500"/> </div><br>
|
||
|
||
```java
|
||
public ListNode EntryNodeOfLoop(ListNode pHead) {
|
||
if (pHead == null || pHead.next == null)
|
||
return null;
|
||
ListNode slow = pHead, fast = pHead;
|
||
do {
|
||
fast = fast.next.next;
|
||
slow = slow.next;
|
||
} while (slow != fast);
|
||
fast = pHead;
|
||
while (slow != fast) {
|
||
slow = slow.next;
|
||
fast = fast.next;
|
||
}
|
||
return slow;
|
||
}
|
||
```
|
||
|
||
# 24. 反转链表
|
||
|
||
[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)
|
||
|
||
## 解题思路
|
||
|
||
### 递归
|
||
|
||
```java
|
||
public ListNode ReverseList(ListNode head) {
|
||
if (head == null || head.next == null)
|
||
return head;
|
||
ListNode next = head.next;
|
||
head.next = null;
|
||
ListNode newHead = ReverseList(next);
|
||
next.next = head;
|
||
return newHead;
|
||
}
|
||
```
|
||
|
||
### 迭代
|
||
|
||
使用头插法。
|
||
|
||
```java
|
||
public ListNode ReverseList(ListNode head) {
|
||
ListNode newList = new ListNode(-1);
|
||
while (head != null) {
|
||
ListNode next = head.next;
|
||
head.next = newList.next;
|
||
newList.next = head;
|
||
head = next;
|
||
}
|
||
return newList.next;
|
||
}
|
||
```
|
||
|
||
# 25. 合并两个排序的链表
|
||
|
||
[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)
|
||
|
||
## 题目描述
|
||
|
||
<div align="center"> <img src="pics/c094d2bc-ec75-444b-af77-d369dfb6b3b4.png" width="400"/> </div><br>
|
||
|
||
## 解题思路
|
||
|
||
### 递归
|
||
|
||
```java
|
||
public ListNode Merge(ListNode list1, ListNode list2) {
|
||
if (list1 == null)
|
||
return list2;
|
||
if (list2 == null)
|
||
return list1;
|
||
if (list1.val <= list2.val) {
|
||
list1.next = Merge(list1.next, list2);
|
||
return list1;
|
||
} else {
|
||
list2.next = Merge(list1, list2.next);
|
||
return list2;
|
||
}
|
||
}
|
||
```
|
||
|
||
### 迭代
|
||
|
||
```java
|
||
public ListNode Merge(ListNode list1, ListNode list2) {
|
||
ListNode head = new ListNode(-1);
|
||
ListNode cur = head;
|
||
while (list1 != null && list2 != null) {
|
||
if (list1.val <= list2.val) {
|
||
cur.next = list1;
|
||
list1 = list1.next;
|
||
} else {
|
||
cur.next = list2;
|
||
list2 = list2.next;
|
||
}
|
||
cur = cur.next;
|
||
}
|
||
if (list1 != null)
|
||
cur.next = list1;
|
||
if (list2 != null)
|
||
cur.next = list2;
|
||
return head.next;
|
||
}
|
||
```
|
||
|
||
# 26. 树的子结构
|
||
|
||
[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)
|
||
|
||
## 题目描述
|
||
|
||
<div align="center"> <img src="pics/84a5b15a-86c5-4d8e-9439-d9fd5a4699a1.jpg" width="450"/> </div><br>
|
||
|
||
## 解题思路
|
||
|
||
```java
|
||
public boolean HasSubtree(TreeNode root1, TreeNode root2) {
|
||
if (root1 == null || root2 == null)
|
||
return false;
|
||
return isSubtreeWithRoot(root1, root2) || HasSubtree(root1.left, root2) || HasSubtree(root1.right, root2);
|
||
}
|
||
|
||
private boolean isSubtreeWithRoot(TreeNode root1, TreeNode root2) {
|
||
if (root2 == null)
|
||
return true;
|
||
if (root1 == null)
|
||
return false;
|
||
if (root1.val != root2.val)
|
||
return false;
|
||
return isSubtreeWithRoot(root1.left, root2.left) && isSubtreeWithRoot(root1.right, root2.right);
|
||
}
|
||
```
|
||
|
||
# 27. 二叉树的镜像
|
||
|
||
[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)
|
||
|
||
## 题目描述
|
||
|
||
<div align="center"> <img src="pics/0c12221f-729e-4c22-b0ba-0dfc909f8adf.jpg" width="300"/> </div><br>
|
||
|
||
## 解题思路
|
||
|
||
```java
|
||
public void Mirror(TreeNode root) {
|
||
if (root == null)
|
||
return;
|
||
swap(root);
|
||
Mirror(root.left);
|
||
Mirror(root.right);
|
||
}
|
||
|
||
private void swap(TreeNode root) {
|
||
TreeNode t = root.left;
|
||
root.left = root.right;
|
||
root.right = t;
|
||
}
|
||
```
|
||
|
||
# 28 对称的二叉树
|
||
|
||
[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)
|
||
|
||
## 题目描述
|
||
|
||
<div align="center"> <img src="pics/0c12221f-729e-4c22-b0ba-0dfc909f8adf.jpg" width="300"/> </div><br>
|
||
|
||
## 解题思路
|
||
|
||
```java
|
||
boolean isSymmetrical(TreeNode pRoot) {
|
||
if (pRoot == null)
|
||
return true;
|
||
return isSymmetrical(pRoot.left, pRoot.right);
|
||
}
|
||
|
||
boolean isSymmetrical(TreeNode t1, TreeNode t2) {
|
||
if (t1 == null && t2 == null)
|
||
return true;
|
||
if (t1 == null || t2 == null)
|
||
return false;
|
||
if (t1.val != t2.val)
|
||
return false;
|
||
return isSymmetrical(t1.left, t2.right) && isSymmetrical(t1.right, t2.left);
|
||
}
|
||
```
|
||
|
||
# 29. 顺时针打印矩阵
|
||
|
||
[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)
|
||
|
||
## 题目描述
|
||
|
||
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
|
||
|
||
<div align="center"> <img src="pics/48517227-324c-4664-bd26-a2d2cffe2bfe.png" width="200px"> </div><br>
|
||
|
||
## 解题思路
|
||
|
||
```java
|
||
public ArrayList<Integer> printMatrix(int[][] matrix) {
|
||
ArrayList<Integer> ret = new ArrayList<>();
|
||
int r1 = 0, r2 = matrix.length - 1, c1 = 0, c2 = matrix[0].length - 1;
|
||
while (r1 <= r2 && c1 <= c2) {
|
||
for (int i = c1; i <= c2; i++)
|
||
ret.add(matrix[r1][i]);
|
||
for (int i = r1 + 1; i <= r2; i++)
|
||
ret.add(matrix[i][c2]);
|
||
if (r1 != r2)
|
||
for (int i = c2 - 1; i >= c1; i--)
|
||
ret.add(matrix[r2][i]);
|
||
if (c1 != c2)
|
||
for (int i = r2 - 1; i > r1; i--)
|
||
ret.add(matrix[i][c1]);
|
||
r1++; r2--; c1++; c2--;
|
||
}
|
||
return ret;
|
||
}
|
||
```
|
||
|
||
|
||
|
||
|
||
</br><div align="center">💡 </br></br> 更多精彩内容将发布在公众号 **CyC2018**，公众号提供了该项目的离线阅读版本，后台回复"下载" 即可领取。也提供了一份技术面试复习思维导图，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"资料" 即可领取。我基本是按照这个思维导图来进行复习的，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思维导图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。</div></br>
|
||
<div align="center"><img width="180px" src="https://cyc-1256109796.cos.ap-guangzhou.myqcloud.com/%E5%85%AC%E4%BC%97%E5%8F%B7.jpg"></img></div>
|