Meta 电话面试经验分享：高效解决算法问题 | 面试代面, OA代做, 面试辅导

Meta 电话面试经验分享：高效解决算法问题 | 面试代面, OA代做, 面试辅导

在这次Meta的电话面试中，面试官给出了两道典型的算法题目，分别涉及到字符串处理和模式匹配。这些问题不仅考察了考生的算法基础，还要求他们在压力下快速做出正确的决策。面对这样的挑战，成功的考生能够在短时间内理清思路，并以优雅的代码实现复杂的问题。

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题目一：Palindrome Validation（回文验证，字符串处理）

Problem Description:

Write a method that takes a string as input and returns true if the string is a palindrome when ignoring punctuation, spaces, and casing. Return false if not. Please determine this in-place.

Example:

- "Race car" -> True
- "A man, a plan, a canal, Panama!" -> True
- "Rats live on no evil star" -> True
- "Hello, world!" -> False

Code:

public class PalindromeValidator {

    public boolean isValid(char c) {
        return Character.isLetterOrDigit(c);
    }

    public char toLowerCase(char c) {
        return Character.toLowerCase(c);
    }

    public boolean isPalindrome(String input) {
        int i = 0, j = input.length() - 1;
        while (i < j) {
            while (i < j && !isValid(input.charAt(i))) {
                i++;
            }
            while (j > i && !isValid(input.charAt(j))) {
                j--;
            }
            if (i < j && toLowerCase(input.charAt(i)) != toLowerCase(input.charAt(j))) {
                return false;
            }
            i++;
            j--;
        }
        return true;
    }

    public static void main(String[] args) {
        PalindromeValidator validator = new PalindromeValidator();
        System.out.println(validator.isPalindrome("Race car")); // True
        System.out.println(validator.isPalindrome("A man, a plan, a canal, Panama!")); // True
        System.out.println(validator.isPalindrome("Rats live on no evil star")); // True
        System.out.println(validator.isPalindrome("Hello, world!")); // False
    }

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题目二：Pattern Matching with Abbreviation（模式匹配，字符串处理）

Problem Description:

Given a pattern string and a target string, check whether or not it’s possible to match the target by substituting the corresponding number of characters for each number in the pattern.

Example:

- Input: "i18n", "internationalization" -> True
- Input: "f6b", "facebook" -> False

Code:

public class PatternMatcher {

    public boolean isDigit(char c) {
        return Character.isDigit(c);
    }

    public int parseInt(String pattern, int[] index) {
        int num = 0;
        while (index[0] < pattern.length() && isDigit(pattern.charAt(index[0]))) {
            num = num * 10 + (pattern.charAt(index[0]) - '0');
            index[0]++;
        }
        return num;
    }

    public boolean canMatch(String pattern, String target) {
        int i = 0, j = 0;
        while (i < pattern.length() && j < target.length()) {
            if (isDigit(pattern.charAt(i))) {
                int[] index = {i};
                int num = parseInt(pattern, index);
                i = index[0];
                j += num;
            } else {
                if (pattern.charAt(i) != target.charAt(j)) {
                    return false;
                }
                i++;
                j++;
            }
        }
        return i == pattern.length() && j == target.length();
    }

    public static void main(String[] args) {
        PatternMatcher matcher = new PatternMatcher();
        System.out.println(matcher.canMatch("i18n", "internationalization")); // True
        System.out.println(matcher.canMatch("f6b", "facebook")); // False
    }