TMCB其他备选电话面试题

前一篇文章是我做的题 这是剩余的三道，都很基础。

1. Write a method to generate the nth number of the Fibonacci sequence.

Defined as

f(0)=0

f(1)=1

f(n)=f(n-1)+ f(n-2) where n>1

Sequence is 0 1 1 2 3 5…

基础题，递归或者非递归方式

public class Fib {

    public static int fib(int n) {
        int result;
        if (n <= 1) {
            return n;
        } else {
            result = fib(n - 2) + fib(n - 1);
        }
        return result;
    }

    public static int getNthFib(int n) {
        if (n <= 1)
            return n; // error & default input
        int fn2 = 1, fn1 = 1, index = 2; // starts from 2
        while (index < n) { // fn2 means f(n-2), fn1 means f(n-1)
            fn1 = fn1 + fn2;
            fn2 = fn1 - fn2;
            index++;
        }
        return fn1;
    }

}

注意fn1和fn2递增的赋值方式，比引入tmp变量更好的解决方案。

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3. Implement an algorithm to find the nth to last element of a singly linked list

findNth(0) would return the last element of the list

findNth(1) would return the second to last element of the list

 

public Node findNthToLast(Node node)

{

 

}

 

基本题，考察两个游标遍历单链表，需要注意边界条件和限制。

public class LastNth {
    public static int findNth(Node head, int n) {
        if (head == null)
            return -1; // error, list is null
        Node fast = head, slow = head;
        int index = 1;
        while (index <= n) {
            if (fast.next != null) {
                fast = fast.next;
            } else {
                return -1; // error, list has less than n elements
            }
            index++;
        }
        while (fast.next != null) {
            fast = fast.next;
            slow = slow.next;
        }
        return slow.val;
    }
}

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4. Write a method to decide if two strings are anagrams or not.

基本题 需要考虑是否是ASCII码（256位）如果是的话那么定义相应位数的数组进行存储

 

public class AnagramStr {

    public static boolean isAnagram(String s1, String s2) {
        int len1 = s1.length(), len2 = s2.length();
        if (len1 != len2) // not same length
            return false;
        int[] letters = new int[256];
        for (int i = 0; i < letters.length; i++) {
            letters[i] = 0;
        }
        for (char c : s1.toCharArray()) {
            ++letters[c];
        }
        for (char c : s2.toCharArray()) {
            --letters[c];
        }
        for (int i = 0; i < letters.length; i++) {
            if (letters[i] != 0) {
                return false;
            }
        }
        return true;
    }
}

 

如果不限于ASCII，可以使用map来对不确定字符数进行统计。

public static boolean areAnagram(String s1, String s2) {
    int len1 = s1.length(), len2 = s2.length();
    HashMap<Character, Integer> map1 = new HashMap<Character, Integer>(), map2 = new HashMap<Character, Integer>();
    if (len1 != len2) // not same length
        return false;
    for (char c : s1.toCharArray()) {
        if (map1.containsKey(c)) {
            map1.put(c, map1.get(c) + 1);
        } else {
            map1.put(c, 1);
        }
    }
    for (char c : s2.toCharArray()) {
        if (map2.containsKey(c)) {
            map2.put(c, map2.get(c) + 1);
        } else {
            map2.put(c, 1);
        }
    }
    if (!map1.equals(map2))
        return false;
    return true;
}

还有一个思路是对两个string进行排序，然后从头往后遍历。