batcoding

JAVA基础代码练习

练习题

Given a string, take the first 2 chars and return the string with the
2 chars added at both the front and back, so "kitten" yields"kikittenki".
If the string length is less than 2, use whatever chars are there.

 **front22("kitten") → "kikittenki"  
   front22("Ha") → "HaHaHa"  
   front22("abc") → "ababcab"**

my code:

 public String front22(String str) {  
     if(str.length() >= 2){
        str = str.substring(0,2) + str + str.substring(0,2);  
     }
     else
        str = str + str + str;
     return str;
 }

solution:

 public String formt22(String str){
	int len = 2;
    if(str.length() < len){
		len = str.length();
	}
	str = str.subString(0 , len) + str + str.subString(0 , len);
 }  

We'll say that a number is "teen" if it is in the range 13..19 inclusive. Given 2 int values, return true if one or the other is teen, but not both.

**loneTeen(13, 99) → true  
loneTeen(21, 19) → true  
loneTeen(13, 13) → false**  

my code:

public boolean loneTeen(int a, int b) {
	if(a >=  13 && a <=19 && (b < 13 || b > 19))
		return true;
	if(b >= 13 && b <= 19 && (a < 13 || a > 19))
		return true;
	return false;
}

sulution:

public boolean longTeen(int a, int b){
	boolean teenA = (a >= 13 && a <= 19)
	booelan teenB = (b > =13 && b <= 19)
	if((teenA && !teenB) || (!teenA && teenB))
		return true;
	return false;
}

Given a string, we'll say that the front is the first 3 chars of the string. If the string length is less than 3, the front is whatever is there. Return a new string which is 3 copies of the front.

 **front3("Java") → "JavJavJav"  
 front3("Chocolate") → "ChoChoCho"  
 front3("abc") → "abcabcabc"**

mycode:

public String front3(String str) {
	if(str.length() < 3)
		return str + str + str;
	else
	   return str.substring(0,3) + str.substring(0,3) + str.substring(0,3);
}

solution:

public String front3(String str) {
    String front;

	if (str.length() >= 3) {
		front = str.substring(0, 3);
	}else {
		front = str;
	}

	return front + front + front;
}

Given a string, return a string made of the first 2 chars (if present), however include first char only if it is 'o' and include the second only if it is 'z', so "ozymandias" yields "oz".

**startOz("ozymandias") → "oz"  
 startOz("bzoo") → "z"  
 startOz("oxx") → "o"**

mycode:

public String startOz(String str) {
	if(str.equals("") || str.equals("o"))
		return str;
	else{
		if(str.length() == 1)	
			return "";
		else{
  			char a = str.charAt(0);
  			char b = str.charAt(1);
  			String str2 = "";
  			if(a == 'o')
    			str2 += "o";
  			if(b == 'z')
    			str2 += "z";
  			return str2;
		} 
	}
}

solution:

public String startOz(String str) {
	String result = "";

	if (str.length() >= 1 && str.charAt(0)=='o') {
		result = result + str.charAt(0);
	}
	
	if (str.length() >= 2 && str.charAt(1)=='z') {
		result = result + str.charAt(1);
	}
	
	return result;
}

Given 2 positive int values, return the larger value that is in the range 10..20 inclusive, or return 0 if neither is in that range.

max1020(11, 19) → 19  
max1020(19, 11) → 19
max1020(11, 9) → 11

mycode:

public int max1020(int a, int b) {
	if( a >= 10 && a <= 20 && b >= 10 && b <=20)
		return a > b ? a : b;
	else if( a >= 10 && a <=20 && (b < 10 || b > 20))
		return a;
	else if((a < 10 || a > 20) && ( b >= 10 && b <= 20))
		return b;
	else 
		return 0;
}

solution:

public int max1020(int a, int b) {
	// First make it so the bigger value is in a
	if (b > a) {
		int temp = a;
		a = b;
		b = temp;
	}
	
	// Knowing a is bigger, just check a first
	if (a >= 10 && a <= 20) return a;
	if (b >= 10 && b <= 20) return b;
	return 0;
}

Given a string and a non-negative int n, we'll say that the front of the string is the first 3 chars, or whatever is there if the string is less than length 3. Return n copies of the front;

 frontTimes("Chocolate", 2) → "ChoCho"  
 frontTimes("Chocolate", 3) → "ChoChoCho"  
 frontTimes("Abc", 3) → "AbcAbcAbc"  

mycode:

public String frontTimes(String str, int n) {
	String font = "";
	String result = "";
	
	if(str.length() <= 3)
		font= str;
	else
		font = str.substring(0,3);
	
	for(int  i = 0 ; i < n ; i++){
		result += font;
	}

	return result;
}

solution:

public String frontTimes(String str, int n) {
	int frontLen = 3;
	
	if (frontLen > str.length()) {
		frontLen = str.length();
	}

	String front = str.substring(0, frontLen);
	String result = "";
	
	for (int i=0; i<n; i++) {
		result = result + front;
	}
	
	return result;
}

Given an array of ints, return true if the sequence of numbers 1, 2, 3 appears in the array somewhere.

 array123([1, 1, 2, 3, 1]) → true    
 array123([1, 1, 2, 4, 1]) → false  
 array123([1, 1, 2, 1, 2, 3]) → true  
 
 这道题基本上是我读错提了，愿意是出现连续的1，2，3,我把他理解成了，只要数组中存在元
 素1，2，3就符合条件，针对这种理解方式，给出了下面的解题方式之一：  

mycode:

public static boolean array123(int[] nums) {  
	if(nums == null || nums.length < 1)
		return false;
	    
	int max = nums[0];  //获取最大值
	for(int  i = 1 ; i < nums.length ; i++){
	  if(nums[i] > max)
      max = nums[i];
	}
	  
	if(max < 3)   //最大数小于3时，一定不满足条件
		return false;
	  
	int [] array = new int [max + 1];  //把数组的值与下标互换，如array[10] = 99 , 改成 array[99] == 1，表示存在99这个数 
	for(int j = 0 ; j < nums.length ; j++){
		array[nums[j]] = 1;
	}
	  
	if(array[1] == 1 && array[2] == 1 && array[3] == 1)
		return true;
	else
	    return false;
}

Given a string, if the first or last chars are 'x', return the string without those 'x' chars, and otherwise return the string unchanged.

 withoutX("xHix") → "Hi"   
 withoutX("xHi") → "Hi"  
 withoutX("Hxix") → "Hxi"  

mycode:

public String withoutX(String str) {
	int place = str.length();
	if(place == 0)
		return str;
	if(str.equals("x"))
		return "";
	else{
		if(str.substring(0,1).equals("x"))
		str= str.substring(1);
	if(str.substring(str.length()-1).equals("x"))
		str = str.substring(0 , str.length()-1);
		return str;
	}
}

solution:

public String withoutX(String str) {
	if (str.length() > 0 && str.charAt(0) == 'x') {
		str = str.substring(1);
	}

	if (str.length() > 0 && str.charAt(str.length()-1) == 'x') {
		str = str.substring(0, str.length()-1);
	}

	return str;

	// Solution  notes: check for the 'x' in both spots. If found, use substring()
	// to grab the part without the 'x'. Check that the length is greater than 0
	// each time -- the need for the second length check is tricky to see.
	// One could .substring() instead of .charAt() to look into the string.

}

知识点积累

1.String字符串的大小写转换

str.toUpperCase();	return String
str.toLowerCase();

2.String已固定字符串传开始/结束是否与指定字符串匹配

str.startsWith(String);   return boolean
str.endsWith(String);  

都有个s不要漏掉了

3.获取String上指定位置的char

str.charAt(int);	return char;

3.替换String中的内容

str.replace(String,String); return String;
str.replaceAll(reg , String);

replace中第一个参数是字符串，replaceAll中第一个参数是正则表达式，他们都是替换字符串中全部的子串

4.查找String中子串的位置

str.indexOf(String);	return int;   //返回第一个子串开始的位置
str.indexOf(String, int);	//从指定位置开始查找第一个子串的位置
str.lastIndexOf(String);	//从后开始查找第一个子串出现的位置
str.lastIndexOf(String , int);//从后开始指定位置前第一个子串出现的位置

5.获取两个数

Math.min(int a, int b)  //获取较小的数
Math.max(int a, int b)  //获取较大的数

6.字符串截取

str.substring(int , int)  
str.substring(int)

第一个参数表示的是字符在字符串中的位置，且范围为 0 - str.length，如果取最大值str.length,则表示什么都不截取到，而不是报数组越界的异常