mirror of
https://github.com/Snailclimb/JavaGuide
synced 2025-06-25 02:27:10 +08:00
fix typo
This commit is contained in:
TommyMerlin
parent
0c61be453d
commit
4a255c9aa3
@ -14,38 +14,36 @@ n<=39
|
||||
**采用迭代法：**
|
||||
|
||||
```java
|
||||
int Fibonacci(int number) {
|
||||
if (number <= 0) {
|
||||
return 0;
|
||||
}
|
||||
if (number == 1 || number == 2) {
|
||||
return 1;
|
||||
}
|
||||
int first = 1, second = 1, third = 0;
|
||||
for (int i = 3; i <= number; i++) {
|
||||
third = first + second;
|
||||
first = second;
|
||||
second = third;
|
||||
}
|
||||
return third;
|
||||
}
|
||||
int Fibonacci(int number) {
|
||||
if (number <= 0) {
|
||||
return 0;
|
||||
}
|
||||
if (number == 1 || number == 2) {
|
||||
return 1;
|
||||
}
|
||||
int first = 1, second = 1, third = 0;
|
||||
for (int i = 3; i <= number; i++) {
|
||||
third = first + second;
|
||||
first = second;
|
||||
second = third;
|
||||
}
|
||||
return third;
|
||||
}
|
||||
```
|
||||
|
||||
**采用递归：**
|
||||
|
||||
```java
|
||||
public int Fibonacci(int n) {
|
||||
|
||||
if (n <= 0) {
|
||||
return 0;
|
||||
}
|
||||
if (n == 1||n==2) {
|
||||
return 1;
|
||||
}
|
||||
public int Fibonacci(int n) {
|
||||
if (n <= 0) {
|
||||
return 0;
|
||||
}
|
||||
if (n == 1||n==2) {
|
||||
return 1;
|
||||
}
|
||||
|
||||
return Fibonacci(n - 2) + Fibonacci(n - 1);
|
||||
|
||||
}
|
||||
return Fibonacci(n - 2) + Fibonacci(n - 1);
|
||||
}
|
||||
```
|
||||
|
||||
### 二 跳台阶问题
|
||||
@ -71,24 +69,24 @@ f(1) = 1, f(2) = 2, f(3) = 3, f(4) = 5， 可以总结出f(n) = f(n-1) + f(n-2)
|
||||
#### **示例代码：**
|
||||
|
||||
```java
|
||||
int jumpFloor(int number) {
|
||||
if (number <= 0) {
|
||||
return 0;
|
||||
}
|
||||
if (number == 1) {
|
||||
return 1;
|
||||
}
|
||||
if (number == 2) {
|
||||
return 2;
|
||||
}
|
||||
int first = 1, second = 2, third = 0;
|
||||
for (int i = 3; i <= number; i++) {
|
||||
third = first + second;
|
||||
first = second;
|
||||
second = third;
|
||||
}
|
||||
return third;
|
||||
}
|
||||
int jumpFloor(int number) {
|
||||
if (number <= 0) {
|
||||
return 0;
|
||||
}
|
||||
if (number == 1) {
|
||||
return 1;
|
||||
}
|
||||
if (number == 2) {
|
||||
return 2;
|
||||
}
|
||||
int first = 1, second = 2, third = 0;
|
||||
for (int i = 3; i <= number; i++) {
|
||||
third = first + second;
|
||||
first = second;
|
||||
second = third;
|
||||
}
|
||||
return third;
|
||||
}
|
||||
```
|
||||
|
||||
### 三 变态跳台阶问题
|
||||
@ -113,9 +111,9 @@ f(n)=f(n-1)+f(n-2)+...+f(1)
|
||||
#### **示例代码：**
|
||||
|
||||
```java
|
||||
int JumpFloorII(int number) {
|
||||
return 1 << --number;//2^(number-1)用位移操作进行，更快
|
||||
}
|
||||
int JumpFloorII(int number) {
|
||||
return 1 << --number;//2^(number-1)用位移操作进行，更快
|
||||
}
|
||||
```
|
||||
|
||||
#### **补充：**
|
||||
@ -124,7 +122,7 @@ f(n)=f(n-1)+f(n-2)+...+f(1)
|
||||
|
||||
1. “<<” : **左移运算符**，等同于乘2的n次方
|
||||
2. “>>”: **右移运算符**，等同于除2的n次方
|
||||
3. “>>>” **无符号右移运算符**，不管移动前最高位是0还是1，右移后左侧产生的空位部分都以0来填充。与>>类似。
|
||||
3. “>>>” : **无符号右移运算符**，不管移动前最高位是0还是1，右移后左侧产生的空位部分都以0来填充。与>>类似。
|
||||
例：
|
||||
int a = 16;
|
||||
int b = a << 2;//左移2，等同于16 * 2的2次方，也就是16 * 4
|
||||
@ -147,22 +145,22 @@ f(n)=f(n-1)+f(n-2)+...+f(1)
|
||||
#### **示例代码：**
|
||||
|
||||
```java
|
||||
public boolean Find(int target, int [][] array) {
|
||||
//基本思路从左下角开始找，这样速度最快
|
||||
int row = array.length-1;//行
|
||||
int column = 0;//列
|
||||
//当行数大于0，当前列数小于总列数时循环条件成立
|
||||
while((row >= 0)&& (column< array[0].length)){
|
||||
if(array[row][column] > target){
|
||||
row--;
|
||||
}else if(array[row][column] < target){
|
||||
column++;
|
||||
}else{
|
||||
return true;
|
||||
}
|
||||
public boolean Find(int target, int [][] array) {
|
||||
//基本思路从左下角开始找，这样速度最快
|
||||
int row = array.length-1;//行
|
||||
int column = 0;//列
|
||||
//当行数大于0，当前列数小于总列数时循环条件成立
|
||||
while((row >= 0)&& (column< array[0].length)){
|
||||
if(array[row][column] > target){
|
||||
row--;
|
||||
}else if(array[row][column] < target){
|
||||
column++;
|
||||
}else{
|
||||
return true;
|
||||
}
|
||||
return false;
|
||||
}
|
||||
return false;
|
||||
}
|
||||
```
|
||||
|
||||
### 五 替换空格
|
||||
@ -175,38 +173,37 @@ f(n)=f(n-1)+f(n-2)+...+f(1)
|
||||
|
||||
这道题不难，我们可以通过循环判断字符串的字符是否为空格，是的话就利用append()方法添加追加“%20”，否则还是追加原字符。
|
||||
|
||||
或者最简单的方法就是利用： replaceAll(String regex,String replacement)方法了，一行代码就可以解决。
|
||||
或者最简单的方法就是利用：replaceAll(String regex,String replacement)方法了，一行代码就可以解决。
|
||||
|
||||
#### **示例代码：**
|
||||
|
||||
**常规做法：**
|
||||
|
||||
```java
|
||||
public String replaceSpace(StringBuffer str) {
|
||||
StringBuffer out=new StringBuffer();
|
||||
for (int i = 0; i < str.toString().length(); i++) {
|
||||
char b=str.charAt(i);
|
||||
if(String.valueOf(b).equals(" ")){
|
||||
out.append("%20");
|
||||
}else{
|
||||
out.append(b);
|
||||
}
|
||||
public String replaceSpace(StringBuffer str) {
|
||||
StringBuffer out = new StringBuffer();
|
||||
for (int i = 0; i < str.toString().length(); i++) {
|
||||
char b = str.charAt(i);
|
||||
if(String.valueOf(b).equals(" ")){
|
||||
out.append("%20");
|
||||
}else{
|
||||
out.append(b);
|
||||
}
|
||||
return out.toString();
|
||||
}
|
||||
return out.toString();
|
||||
}
|
||||
```
|
||||
|
||||
**一行代码解决：**
|
||||
|
||||
```java
|
||||
public String replaceSpace(StringBuffer str) {
|
||||
//return str.toString().replaceAll(" ", "%20");
|
||||
//public String replaceAll(String regex,String replacement)
|
||||
//用给定的替换替换与给定的regular expression匹配的此字符串的每个子字符串。
|
||||
//\ 转义字符. 如果你要使用 "\" 本身, 则应该使用 "\\". String类型中的空格用“\s”表示，所以我这里猜测"\\s"就是代表空格的意思
|
||||
return str.toString().replaceAll("\\s", "%20");
|
||||
}
|
||||
|
||||
public String replaceSpace(StringBuffer str) {
|
||||
//return str.toString().replaceAll(" ", "%20");
|
||||
//public String replaceAll(String regex,String replacement)
|
||||
//用给定的替换替换与给定的regular expression匹配的此字符串的每个子字符串。
|
||||
//\ 转义字符. 如果你要使用 "\" 本身, 则应该使用 "\\". String类型中的空格用“\s”表示，所以我这里猜测"\\s"就是代表空格的意思
|
||||
return str.toString().replaceAll("\\s", "%20");
|
||||
}
|
||||
```
|
||||
|
||||
### 六 数值的整数次方
|
||||
@ -279,17 +276,17 @@ public class Solution {
|
||||
当然这一题也可以采用笨方法：累乘。不过这种方法的时间复杂度为O（n），这样没有前一种方法效率高。
|
||||
|
||||
```java
|
||||
// 使用累乘
|
||||
public double powerAnother(double base, int exponent) {
|
||||
double result = 1.0;
|
||||
for (int i = 0; i < Math.abs(exponent); i++) {
|
||||
result *= base;
|
||||
}
|
||||
if (exponent >= 0)
|
||||
return result;
|
||||
else
|
||||
return 1 / result;
|
||||
// 使用累乘
|
||||
public double powerAnother(double base, int exponent) {
|
||||
double result = 1.0;
|
||||
for (int i = 0; i < Math.abs(exponent); i++) {
|
||||
result *= base;
|
||||
}
|
||||
if (exponent >= 0)
|
||||
return result;
|
||||
else
|
||||
return 1 / result;
|
||||
}
|
||||
```
|
||||
|
||||
### 七 调整数组顺序使奇数位于偶数前面
|
||||
@ -434,22 +431,21 @@ public class ListNode {
|
||||
}
|
||||
}*/
|
||||
public class Solution {
|
||||
public ListNode ReverseList(ListNode head) {
|
||||
ListNode next = null;
|
||||
ListNode pre = null;
|
||||
while (head != null) {
|
||||
//保存要反转到头来的那个节点
|
||||
next = head.next;
|
||||
//要反转的那个节点指向已经反转的上一个节点
|
||||
head.next = pre;
|
||||
//上一个已经反转到头部的节点
|
||||
pre = head;
|
||||
//一直向链表尾走
|
||||
head = next;
|
||||
public ListNode ReverseList(ListNode head) {
|
||||
ListNode next = null;
|
||||
ListNode pre = null;
|
||||
while (head != null) {
|
||||
//保存要反转到头来的那个节点
|
||||
next = head.next;
|
||||
//要反转的那个节点指向已经反转的上一个节点
|
||||
head.next = pre;
|
||||
//上一个已经反转到头部的节点
|
||||
pre = head;
|
||||
//一直向链表尾走
|
||||
head = next;
|
||||
}
|
||||
return pre;
|
||||
}
|
||||
return pre;
|
||||
}
|
||||
|
||||
}
|
||||
```
|
||||
|
||||
@ -538,20 +534,20 @@ public class Solution {
|
||||
|
||||
```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;
|
||||
}
|
||||
}
|
||||
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;
|
||||
}
|
||||
}
|
||||
```
|
||||
|
||||
### 十一 用两个栈实现队列
|
||||
@ -642,8 +638,6 @@ https://www.nowcoder.com/questionTerminal/d77d11405cc7470d82554cb392585106
|
||||
….
|
||||
依次执行，最后辅助栈为空。如果不为空说明弹出序列不是该栈的弹出顺序。
|
||||
|
||||
|
||||
|
||||
#### **考察内容：**
|
||||
|
||||
栈
|
||||
|
||||
Loading…
x
Reference in New Issue
Block a user