第二章模拟算法和字符串处理技巧

分类: IT文章 • 2023-11-25 09:08:13

1 Sliding Window Average from Data Stream

public class MovingAvage {

    int id, size;
    double[] sum;
    
    public MovingAvage(int s) {
        size = s;
        sum = new double[100000];
    }
    
    public double add(int num) {
        id++;
        sum[id] = sum[id - 1] + num;
        if (id - size >= 0) {
            return (sum[id] - sum[id - size]) / size;
        } else {
            return sum[id] / id;
        }
    }
}

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使用滚动数组 1 缩减数组 2 写mod函数 3 所有用到数组的地方改变量

public class MovingAvage {

    int id, size;
    double[] sum;
    
    public MovingAvage(int s) {
        size = s;
        sum = new double[size + 1]; // 1
    }
    
    int mod(int x) {  // 2
        return x % (size + 1);
    }
    // 3
    public double add(int num) {
        id++;
        sum[mod(id)] = sum[mod(id - 1)] + num;
        if (id - size >= 0) {
            return (sum[mod(id)] - sum[mod(id - size)]) / size;
        } else {
            return sum[mod(id)] / id;
        }
    }
}

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2 Edit Distance II

    public boolean isOneEditDistance(String s, String t) {
        // write your code here
        if (s.length() > t.length()) {
            return isOneEditDistance(t, s);
        }
        
        int diff = t.length() - s.length();
        if (diff > 1) {
            return false;
        }
        
        if (diff == 0) {
            int con = 0;
            for (int i = 0; i < s.length(); i++) {
                if (s.charAt(i) != t.charAt(i)) {
                    con++;
                }
            }
            return con == 1;
        }
        if (diff == 1) {
            for (int i = 0; i < s.length(); i++) {
                if (s.charAt(i) != t.charAt(i)) {
                    return s.substring(i).equals(t.substring(i + 1));
                }
            }
        }
        return true;
    }

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3 Read Characters From File - multiple calls

public class MovingAvage {

    char[] buffer = new char[4];
    int head, tail;
    
    public int read(char[] buf, int n) {
        int i = 0;
        while (i < n) {
            if (head == tail) {
                head = 0;
                tail = read4(buffer);
                if (tail == 0) {
                    break;
                }
            }
            while (i < n && head < tail) {
                buf[i++] = buffer[head++];
            }
        }
        return i;
    }
}

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4 Strings Serialization

public class MovingAvage {

    String encode(List<String> list) {
        StringBuilder sb = new StringBuilder();
        for (String s : list) {
            for (char c : s.toCharArray()) {
                if (c == ':') {
                    sb.append("::");
                } else {
                    sb.append(c);
                }
            }
            sb.append(":;");
        }
        return sb.toString();
    }
    
    List<String> decode(String str) {
        List<String> res = new ArrayList<>();
        char[] sc = str.toCharArray();
        StringBuilder item = new StringBuilder();
        int i = 0;
        while (i < sc.length) {
            if (sc[i] == ':') {
                if (sc[i + 1] == ';') {
                    res.add(item.toString());
                    item = new StringBuilder();
                    i += 2;
                } else {
                    item.append(sc[i + 1]);
                    i += 2;
                }
            } else {
                item.append(sc[i]);
                i++;
            }
        }
        return res;
    }
}

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5 System Longest File Path

    public int lengthLongestPath(String input) {
        // write your code here
        if (input.length() == 0) {
            return 0;
        }
        
        int[] level = new int[input.length() + 1];
        
        int res = 0;
        for (String line : input.split("
")) {
            int l = line.lastIndexOf("	") + 2;
            int len = line.length() - (l - 1);
            if (line.contains(".")) {
                res = Math.max(res, level[l - 1] + len);
            } else {
                level[l] = level[l - 1] + len + 1;
            }
        }
        return res;
    }

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6 Roman to Integer

    public int romanToInt(String s) 
    {
        if (s.length() == 0) {
            return 0;
        }
         char[] cs = s.toCharArray();
        int res = toInt(cs[0]);
        for (int i = 1; i < cs.length; i++) {
            res += toInt(cs[i]);
            if (toInt(cs[i - 1]) < toInt(cs[i])) {
                res -= toInt(cs[i - 1]) * 2;
            }
        }
        return res;
    }
    int toInt(char c) {
        switch(c) {
            case 'I': return 1;
            case 'V': return 5;
            case 'X': return 10;
            case 'L': return 50;
            case 'C': return 100;
            case 'D': return 500;
            case 'M': return 1000;
        }
        return 0;
    }

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7 Integer to Roman

public static String intToRoman(int num) {
    String M[] = {"", "M", "MM", "MMM"};
    String C[] = {"", "C", "CC", "CCC", "CD", "D", "DC", "DCC", "DCCC", "CM"};
    String X[] = {"", "X", "XX", "XXX", "XL", "L", "LX", "LXX", "LXXX", "XC"};
    String I[] = {"", "I", "II", "III", "IV", "V", "VI", "VII", "VIII", "IX"};
    return M[num/1000] + C[(num%1000)/100] + X[(num%100)/10] + I[num%10];
}

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8 identify celebrity

    int findCelebrity(int n) {
        int res = 0;
        for (int i = 1; i < n; i++) {
            if (know(res, i)) {
                res = i;
            }
        }
        for (int i = 0; i < n; i++) {
            if (res != i && know(res, i)) {
                return -1;
            }
            if (res != i && !know(i, res)) {
                return -1;
            }
        }
        return res;
    }

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第二章 模拟算法和字符串处理技巧

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第二章模拟算法和字符串处理技巧