字符串类

leetcode 49. 字母异位词分组

方法一：将字符串按字符排序好作为key

public List<List<String>> groupAnagrams(String[] strs) {
        if (strs == null || strs.length == 0) {
            return new ArrayList<>();
        }
        Map<String, List<String>> map = new HashMap<>();
        for (String str : strs) {
            char[] chars = str.toCharArray();
            Arrays.sort(chars);
            String key = String.valueOf(chars);
            if (map.containsKey(key)) {
                map.get(key).add(str);
            } else {
                List<String> list = new ArrayList<>();
                list.add(str);
                map.put(key, list);
            }
        }
        return new ArrayList<List<String>>(map.values());
    }

方法二：统计字符串每个字符出现的次数作为可以

public List<List<String>> groupAnagrams(String[] strs) {
        if (strs == null || strs.length == 0) {
            return new ArrayList<>();
        }
        Map<String, List<String>> map = new HashMap<>();
        for (String str : strs) {
            int[] table = new int[26];
            for (int i = 0; i < str.length(); i++) {
                table[str.charAt(i) - 'a']++;
            }
            StringBuffer keySb = new StringBuffer();
            for (int i = 0; i < 26; i++) {
                keySb.append(table[i] + "#");
            }
            String key = keySb.toString();
            if (map.containsKey(key)) {
                map.get(key).add(str);
            } else {
                List<String> list = new ArrayList<>();
                list.add(str);
                map.put(key, list);
            }
        }
        return new ArrayList<>(map.values());

    }

leetcode 58. 最后一个单词的长度

public int lengthOfLastWord(String s) {
        int count = 0;
        int index = s.length() - 1;
        //过滤空格
        while (index >= 0 && s.charAt(index) == ' ') {
            index--;
        }
        //计算最后一个单词的长度
        for (int i = index; i >= 0; i--) {
            if (s.charAt(i) == ' ') {
                break;
            }
            count++;
        }
        return count;
    }

leetcode 67. 二进制求和

提前申请好空间，避免后面需要重新申请

public String addBinary(String a, String b) {
        if (a == null || a.length() == 0) {
            return b;
        }
        if (b == null || b.length() == 0) {
            return a;
        }
        if (a.length() < b.length()) {
            return addBinary(b, a);
        }
        int len = a.length() + 1;
        char[] newCharArray = new char[len];
        int flag = 0;
        int num = 0;
        int gap = a.length() - b.length();
        for (int i = a.length() - 1; i >= gap; i--) {
            num = flag + a.charAt(i) + b.charAt(i - gap) - '0' * 2;
            newCharArray[--len] = num % 2 == 0 ? '0' : '1';
            flag = num / 2;
        }
        for (int i = gap - 1; i >= 0; i--) {
            num = flag + a.charAt(i) - '0';
            newCharArray[--len] = num % 2 == 0 ? '0' : '1';
            flag = num / 2;
        }
        if (flag == 1) {
            newCharArray[0] = '1';
        }
        if (newCharArray[0] == '1') {
            return new String(newCharArray);
        }
        return new String(newCharArray, 1, a.length());

    }