今天题不难，写个题解吧

1 万人千题，跟着英雄哥有肉吃

https://blog.youkuaiyun.com/WhereIsHeroFrom/article/details/121260397

2 打卡第22讲

《算法零基础100讲》(第22讲) 字符串算法(二) - 字符串比较

https://blog.youkuaiyun.com/WhereIsHeroFrom/article/details/120875787

3 题目

3.1 剑指 Offer 05. 替换空格

如此水题解，有些羞愧

class Solution {
    public String replaceSpace(String s) {
        return s.replaceAll(" ","%20");
    }
}

3.2 面试题 10.05. 稀疏数组搜索

二分查找还是不熟练，需要练练

class Solution {
    public int findString(String[] words, String s) {
        return search(words, s, 0, words.length-1);
    }

    public int search(String[] words, String s, int l, int r) {
        if (l >= r) {
            if (words[l].equals(s)) {
                return l;
            } else {
                return -1;
            }
        }

        int mid = (r + l) / 2;
        int index = mid;
        while (index > 0 && words[index].length() == 0) {
            index--;
        }

        if (index == 0 && words[index].length() == 0) {
            return search(words, s, mid + 1, r);
        }

        if (words[index].equals(s)) {
            return index;
        }

        if (words[index].compareTo(s) > 0) {
            return search(words, s, l, mid - 1);
        } else if (words[index].compareTo(s) < 0) {
            return search(words, s, mid + 1, r);
        }
        return -1;
    }
}

3.3 290. 单词规律

class Solution {
    public boolean wordPattern(String pattern, String s) {
        String[] words = s.split(" ");
        if (words.length != pattern.length()) {
            return false;
        }

        Map<Character, String> map = new HashMap<>();
        Map<String, Character> map2 = new HashMap<>();
        for (int i = 0; i < words.length; i++) {
            char c = pattern.charAt(i);
            String record = map.get(c);
            Character p = map2.get(words[i]);
            if (record == null && p == null) {
                map.put(c, words[i]);
                map2.put(words[i], c);
                continue;
            }

            if (p == null || p != c) {
                return false;
            }

            if (!words[i].equals(map.get(c))) {
                return false;
            }
        }

        return true;
    }
}

3.4 1309. 解码字母到整数映射

public String freqAlphabets(String s) {
        char[] chars = {'0', 'a', 'b', 'c', 'd', 'e',
                'f', 'g', 'h', 'i', 'j',
                'k', 'l', 'm', 'n', 'o',
                'p', 'q', 'r', 's', 't',
                'u', 'v', 'w', 'x', 'y',
                'z'
        };
        StringBuilder sb = new StringBuilder();
        int i = 0;
        while (i < s.length()) {
            if (i >= s.length()) {
                break;
            }

            if (i + 2 < s.length() && s.charAt(i + 2) == '#') {
                char[] charNum = {s.charAt(i), s.charAt(i + 1)};
                int num = Integer.valueOf(String.valueOf(charNum));
                sb.append(chars[num]);
                i += 3;
            } else {
                int num = s.charAt(i) - '1' + 1;
                sb.append(chars[num]);
                i++;
            }
        }
        return sb.toString();
    }

3.5 1967. 作为子字符串出现在单词中的字符串数目

又水一道，逃

class Solution {
    public int numOfStrings(String[] patterns, String word) {
        if (patterns.length == 0) {
            return 0;
        }

        int count = 0;
        for (String p : patterns) {
            if (word.contains(p)) {
                count++;
            }
        }
        return count;
    }
}