笔试中遇到的问题
KMP算法
KMP算法是一种字符串匹配算法。匹配过程详见wiki或者阮一峰的博客。
public class KMP {
private final int R; // the radix
private int[][] dfa; // the KMP automoton
private char[] pattern; // either the character array for the pattern
private String pat; // or the pattern string
/**
* Preprocesses the pattern string.
*
* @param pat the pattern string
*/
public KMP(String pat) {
this.R = 256;
this.pat = pat;
// build DFA from pattern
int m = pat.length();
dfa = new int[R][m];
dfa[pat.charAt(0)][0] = 1;
for (int x = 0, j = 1; j < m; j++) {
for (int c = 0; c < R; c++)
dfa[c][j] = dfa[c][x]; // Copy mismatch cases.
dfa[pat.charAt(j)][j] = j+1; // Set match case.
x = dfa[pat.charAt(j)][x]; // Update restart state.
}
}
/**
* Preprocesses the pattern string.
*
* @param pattern the pattern string
* @param R the alphabet size
*/
public KMP(char[] pattern, int R) {
this.R = R;
this.pattern = new char[pattern.length];
for (int j = 0; j < pattern.length; j++)
this.pattern[j] = pattern[j];
// build DFA from pattern
int m = pattern.length;
dfa = new int[R][m];
dfa[pattern[0]][0] = 1;
for (int x = 0, j = 1; j < m; j++) {
for (int c = 0; c < R; c++)
dfa[c][j] = dfa[c][x]; // Copy mismatch cases.
dfa[pattern[j]][j] = j+1; // Set match case.
x = dfa[pattern[j]][x]; // Update restart state.
}
}
/**
* Returns the index of the first occurrrence of the pattern string
* in the text string.
*
* @param txt the text string
* @return the index of the first occurrence of the pattern string
* in the text string; N if no such match
*/
public int search(String txt) {
// simulate operation of DFA on text
int m = pat.length();
int n = txt.length();
int i, j;
for (i = 0, j = 0; i < n && j < m; i++) {
j = dfa[txt.charAt(i)][j];
}
if (j == m) return i - m; // found
return n; // not found
}
/**
* Returns the index of the first occurrrence of the pattern string
* in the text string.
*
* @param text the text string
* @return the index of the first occurrence of the pattern string
* in the text string; N if no such match
*/
public int search(char[] text) {
// simulate operation of DFA on text
int m = pattern.length;
int n = text.length;
int i, j;
for (i = 0, j = 0; i < n && j < m; i++) {
j = dfa[text[i]][j];
}
if (j == m) return i - m; // found
return n; // not found
}
/**
* Takes a pattern string and an input string as command-line arguments;
* searches for the pattern string in the text string; and prints
* the first occurrence of the pattern string in the text string.
*
* @param args the command-line arguments
*/
public static void main(String[] args) {
String pat = args[0];
String txt = args[1];
char[] pattern = pat.toCharArray();
char[] text = txt.toCharArray();
KMP kmp1 = new KMP(pat);
int offset1 = kmp1.search(txt);
KMP kmp2 = new KMP(pattern, 256);
int offset2 = kmp2.search(text);
// print results
System.out.println("text: " + txt);
System.out.print("pattern: ");
for (int i = 0; i < offset1; i++)
System.out.print(" ");
System.out.println(pat);
System.out.print("pattern: ");
for (int i = 0; i < offset2; i++)
System.out.print(" ");
System.out.println(pat);
}
}二分查找
/**
* 二分查找
* @param sorted
* @param target
* @return
*/
public int binarySearch(int[] sorted, int target) {
int low = 0;
int high = sorted.length - 1;
while (low <= high) {
int mid = (low + high) / 2;
if (sorted[mid] == target) {
return mid;
}
if (sorted[mid] > target) {
high = mid - 1;
}
if (sorted[mid] < target) {
low = mid + 1;
}
}
return -1;
}例如：Array：2 5 8 12 16 23 38 56 72 91，Target：23。则pivot分别是：16 -> 56
CSMA/CD
Carrier-sense multiple access with collision detection (CSMA/CD) is a media access control method used most notably in early Ethernet technology for local area networking.
一项局域网技术。
循环冗余算法
一种数据校验算法，英文缩写是CRC。
局域网技术
- CSMA/CD
- Token ring
- AppleTalk
- ATM
关于IP地址的分配
线程资源
共享的有：
- 进程代码段
- 进程的公有数据
- 进程打开的文件描述符
- 信号的处理器
- 进程的当前目录和进程用户ID与进程组ID
线程私有资源：
- 线程ID
- 寄存器组的值
- 线程的堆栈
- 错误返回码
- 线程的信号屏蔽码
- 线程的优先级
广度优先和深度优先遍历
排列组合
排列是有序的；组合是无序的。
特殊元素（位置）优先安排法
例如：用0,2,3,4,5五个数字，组合没有重复数字的三位数，问偶数有几个？（30）
解：这里的0不能在第一位，所以是特殊元素，先考虑。总体淘汰法（排除法）
相邻捆绑法
解决相邻问题。不相邻插空法
解决不相邻问题。顺序固定用除法
例如：6个人排队，甲乙丙三人按甲乙丙顺序排列的方法有多少种？
解：A66 / A33 = 120分排问题直排法
多排问题转一排。无脑实验法
隔板法
例如：方程a + b + c + d = 12有多少组正整数解？
解：转化为12个球分4份的问题。对偶法
“至多”“至少”情况；比如至少一个，那他的对立面就是都没有。
数组类型
一个数组初始化为什么类型，则接下来存入的元素只能是该类型或子类。
public class TestArrayObject {
public static void main(String[] args) {
A[] as = new B[10];
as[0] = new C(); // would be error
as[1] = new BB();
as[2] = new A(); // would be error
}
}
class A {}
class B extends A {}
class C extends A {}
class BB extends B {}SQL组合索引
ALTER TABLE myIndex ADD INDEX name_city_age(vc_Name(10),vc_City,i_Age);相当于建立了：
vc_Name,vc_City,i_Age
vc_Name,vc_City
vc_Name