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

转载于:https://www.cnblogs.com/jinxiuze/p/7662000.html

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