剑指offer笔记

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面试题1：实现singleton模式

代码实现：

public class Singleton {
private static Singleton instance = null;
private static Object syncObj = new Object();
private Singleton() {
}
public static Singleton getInstance() {
if (instance == null) {
synchronized (syncObj) {
if (instance == null) {
instance = new Singleton();
}
}
}
return instance;
}
}

强烈推荐：静态构造方法

public class Singleton1{
private static Singleton1 instance = new Singleton1();
private Singleton1(){
}
public static Singleton1 getInstance(){
return instance;
}
}

面试题3：二维数组中查找

代码：

public static boolean find(int[][] arr, int rows, int columns, int number) {
boolean found = false;
if (arr == null || rows == 0 || columns == 0) {
return found;
}
int row = 0;
int column = columns - 1;
while (row < rows && column >= 0) {
if (arr[row][column] == number) {
found = true;
return found;
} else if (arr[row][column] > number) {
--column;
} else {
++row;
}
}
return found;
}

面试题4：替换空格

代码：

public static char[] replaceBlank(char[] ch) {
int originLength = 0;
int i = 0;
int numberOfBlank = 0;
while (ch[i] != '\0') {
originLength++;
if (ch[i] == ' ') {
numberOfBlank++;
}

i++;
}

int newLength = originLength + 2 * numberOfBlank;
if (ch.length < newLength) {
return null;
}

int indexOfOrigin = originLength - 1;
int indexOfNew = newLength - 1;
while (indexOfOrigin >= 0 && indexOfNew > indexOfOrigin) {
if (ch[indexOfOrigin] == ' ') {
ch[indexOfNew--] = '0';
ch[indexOfNew--] = '2';
ch[indexOfNew--] = '%';
} else {
ch[indexOfNew--] = ch[indexOfOrigin];
}
indexOfOrigin--;
}
return ch;
}

/**
* 合并两个有序数组A1、A2, A1有足够多的空余空间 O(n)
*/
public static int[] mergeArray(int[] a1,int a2) {
int i = 0;
int a1Length = 0;
while (a1[i++] != '\0') {
a1Length++;
}
a1Length--;
int a2Length = a2.length - 1;
int newLength = a1Length + a2Length + 1;

while (newLength >= 0) {
if (a1Length < 0 && a2Length >= 0) {
a1[newLength--] = a2[a2Length--];
continue;
}
if (a2[a2Length] >= a1[a1Length]) {
a1[newLength--] = a2[a2Length--];
} else {
a1[newLength--] = a1[a1Length--];
}
}
return a1;
}

三、高质量的代码

3.1 代码的完整性

3.1.1 第三种错误处理的方法

面试题11：数值的整数次方

优化：

面试题12：

面试题 13：

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