java笔试编写算法面试题含答案

1.统计一篇英文文章单词个数。

public class WordCounting {
    public static void main(String[] args) {
        try(FileReader fr = new FileReader("a.txt")) {
            int counter = 0;
            boolean state = false;
            int currentChar;
            while((currentChar= fr.read()) != -1) {
                if(currentChar== ' ' || currentChar == '\n'
                        || currentChar == '\t' || currentChar == '\r') {
                    state = false;
                }
                else if(!state) {
                    state = true;
                    counter  ;
                }
            }
            System.out.println(counter);
        }
        catch(Exception e) {
            e.printStackTrace();
        }
    }
}

补充：这个程序可能有很多种写法，这里选择的是Dennis M. Ritchie和Brian W. Kernighan老师在他们不朽的著作《The C Programming Language》中给出的代码，向两位老师致敬。下面的代码也是如此。

2.输入年月日，计算该日期是这一年的第几天。

public class DayCounting {
    public static void main(String[] args) {
        int[][] data = {
                {31,28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31},
                {31,29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}
        };
        Scanner sc = new Scanner(System.in);
        System.out.print("请输入年月日(1980 11 28): ");
        int year = sc.nextInt();
        int month = sc.nextInt();
        int date = sc.nextInt();
        int[] daysOfMonth = data[(year % 4 == 0 && year % 100 != 0 || year % 400 == 0)?1 : 0];
        int sum = 0;
        for(int i = 0; i < month -1; i  ) {
            sum  = daysOfMonth[i];
        }
        sum  = date;
        System.out.println(sum);
        sc.close();
    }
}

3.回文素数：所谓回文数就是顺着读和倒着读一样的数(例如：11，121，1991…)，回文素数就是既是回文数又是素数(只能被1和自身整除的数)的数。编程找出11～9999之间的回文素数。

public class PalindromicPrimeNumber {
    public static void main(String[] args) {
        for(int i = 11; i <= 9999; i  ) {
            if(isPrime(i) && isPalindromic(i)) {
                System.out.println(i);
            }
        }
    }
    public static boolean isPrime(int n) {
        for(int i = 2; i <= Math.sqrt(n); i  ) {
            if(n % i == 0) {
                return false;
            }
        }
        return true;
    }
    public static boolean isPalindromic(int n) {
        int temp = n;
        int sum = 0;
        while(temp > 0) {
            sum= sum * 10   temp % 10;
            temp/= 10;
        }
        return sum == n;
    }
}

4.全排列：给出五个数字12345的所有排列。

public class FullPermutation {
    public static void perm(int[] list) {
        perm(list,0);
    }
    private static void perm(int[] list, int k) {
        if (k == list.length) {
            for (int i = 0; i < list.length; i  ) {
                System.out.print(list[i]);
            }
            System.out.println();
        }else{
            for (int i = k; i < list.length; i  ) {
                swap(list, k, i);
                perm(list, k   1);
                swap(list, k, i);
            }
        }
    }
    private static void swap(int[] list, int pos1, int pos2) {
        int temp = list[pos1];
        list[pos1] = list[pos2];
        list[pos2] = temp;
    }
    public static void main(String[] args) {
        int[] x = {1, 2, 3, 4, 5};
        perm(x);
    }
}

5.对于一个有N个整数元素的一维数组，找出它的子数组（数组中下标连续的元素组成的数组）之和的最大值。

下面给出几个例子（最大子数组用粗体表示）：

数组：{ 1, -2, 3,5, -3, 2 }，结果是：8

2) 数组：{ 0, -2, 3, 5, -1, 2 }，结果是：9

3) 数组：{ -9, -2,-3, -5, -3 }，结果是：-2

可以使用动态规划的思想求解：

public class MaxSum {
    private static int max(int x, int y) {
        return x > y? x: y;
    }
    public static int maxSum(int[] array) {
        int n = array.length;
        int[] start = new int[n];
        int[] all = new int[n];
        all[n - 1] = start[n - 1] = array[n - 1];
        for(int i = n - 2; i >= 0;i--) {
            start[i] = max(array[i], array[i]   start[i   1]);
            all[i] = max(start[i], all[i   1]);
        }
        return all[0];
    }
    public static void main(String[] args) {
        int[] x1 = { 1, -2, 3, 5,-3, 2 };
        int[] x2 = { 0, -2, 3, 5,-1, 2 };
        int[] x3 = { -9, -2, -3,-5, -3 };
        System.out.println(maxSum(x1)); // 8
        System.out.println(maxSum(x2)); // 9
        System.out.println(maxSum(x3)); //-2
    }
}

6.用递归实现字符串倒转

public class StringReverse {
    public static String reverse(String originStr) {
        if(originStr == null || originStr.length()== 1) {
            return originStr;
        }
        return reverse(originStr.substring(1))  originStr.charAt(0);
    }
    public static void main(String[] args) {
        System.out.println(reverse("hello"));
    }
}

7.输入一个正整数，将其分解为素数的乘积。

public class DecomposeInteger {
    private static List<Integer> list = new ArrayList<Integer>();
    public static void main(String[] args) {
        System.out.print("请输入一个数: ");
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        decomposeNumber(n);
        System.out.print(n   " = ");
        for(int i = 0; i < list.size() - 1; i  ) {
            System.out.print(list.get(i)   " * ");
        }
        System.out.println(list.get(list.size() - 1));
    }
    public static void decomposeNumber(int n) {
        if(isPrime(n)) {
            list.add(n);
            list.add(1);
        }
        else {
            doIt(n, (int)Math.sqrt(n));
        }
    }
    public static void doIt(int n, int div) {
        if(isPrime(div) && n % div == 0) {
            list.add(div);
            decomposeNumber(n / div);
        }
        else {
            doIt(n, div - 1);
        }
    }
    public static boolean isPrime(int n) {
        for(int i = 2; i <= Math.sqrt(n);i  ) {
            if(n % i == 0) {
                return false;
            }
        }
        return true;
    }
}

8、一个有n级的台阶，一次可以走1级、2级或3级，问走完n级台阶有多少种走法。

public class GoSteps {
    public static int countWays(int n) {
        if(n < 0) {
            return 0;
        }
        else if(n == 0) {
            return 1;
        }
        else {
            return countWays(n - 1)   countWays(n - 2)   countWays(n -3);
        }
    }
    public static void main(String[] args) {
        System.out.println(countWays(5)); // 13
    }
}

9.写一个算法判断一个英文单词的所有字母是否全都不同（不区分大小写）

public class AllNotTheSame {
    public static boolean judge(String str) {
        String temp = str.toLowerCase();
        int[] letterCounter = new int[26];
        for(int i = 0; i <temp.length(); i  ) {
            int index = temp.charAt(i)- 'a';
            letterCounter[index]  ;
            if(letterCounter[index] > 1) {
                return false;
            }
        }
        return true;
    }
    public static void main(String[] args) {
        System.out.println(judge("hello"));
        System.out.print(judge("smile"));
    }
}

10.有一个已经排好序的整数数组，其中存在重复元素，请将重复元素删除掉，例如，A= [1, 1, 2, 2, 3]，处理之后的数组应当为A= [1, 2, 3]。

public class RemoveDuplication {
    public static int[] removeDuplicates(int a[]) {
        if(a.length <= 1) {
            return a;
        }
        int index = 0;
        for(int i = 1; i < a.length; i  ) {
            if(a[index] != a[i]) {
                a[  index] = a[i];
            }
        }
        int[] b = new int[index   1];
        System.arraycopy(a, 0, b, 0, b.length);
        return b;
    }
    public static void main(String[] args) {
        int[] a = {1, 1, 2, 2, 3};
        a = removeDuplicates(a);
        System.out.println(Arrays.toString(a));
    }
}

11.给一个数组，其中有一个重复元素占半数以上，找出这个元素。

public class FindMost {
    public static <T> T find(T[] x){
        T temp = null;
        for(int i = 0, nTimes = 0; i< x.length;i  ) {
            if(nTimes == 0) {
                temp= x[i];
                nTimes= 1;
            }
            else {
                if(x[i].equals(temp)) {
                    nTimes  ;
                }
                else {
                    nTimes--;
                }
            }
        }
        return temp;
    }
    public static void main(String[] args) {
        String[]strs = {"hello","kiss","hello","hello","maybe"};
        System.out.println(find(strs));
    }
}

12.编写一个方法求一个字符串的字节长度?

public int getWordCount(String s){
    int length = 0;
    for(int i = 0; i < s.length(); i  )
    {
        int ascii = Character.codePointAt(s, i);
        if(ascii >= 0 && ascii <=255)
            length  ;
        else
            length  = 2;
    }
    return length;
}

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