字符串

LC 3

给定一个字符串 s ，请你找出其中不含有重复字符的 最长子串 的长度。

public class LengthOfLongestSubstring {
    public int lengthOfLongestSubstring(String s) {
        if (s.equals(null) || s.length() == 0)
            return 0;
        Map<Character, Integer> map = new HashMap<>();
        int ans = 0;
        for (int start = 0, end = 0; end < s.length(); end++) {
            char alpha = s.charAt(end);
            if (map.containsKey(alpha))
                start = Math.max(map.get(alpha), start);
            ans = Math.max(ans, end - start + 1);
            map.put(s.charAt(end), end + 1);
        }
        return ans;
    }
}

LC 13

罗马数字转整数

// 做法1：
class LC13 {
    public static int romanToInt(String s) {
        int n = s.length();
        int roman_int = 0;
        for(int i = 0; i < n; i++) {
            switch(s.charAt(i)) {
                case 'I':
                    roman_int = roman_int + 1;
                    break;
                case 'V':
                    roman_int = roman_int + 5;
                    break;
                case 'X':
                    roman_int = roman_int + 10;
                    break;
                case 'L':
                    roman_int = roman_int + 50;
                    break;
                case 'C':
                    roman_int = roman_int + 100;
                    break;
                case 'D':
                    roman_int = roman_int + 500;
                    break;
                case 'M':
                    roman_int = roman_int + 1000;
                    break;
                default:
                    System.out.println("default");
                    break;
            }
			
            // 例外判断
            if(i != 0) {
                if(((s.charAt(i) == 'V') || (s.charAt(i) == 'X')) && (s.charAt(i - 1) == 'I'))
                    roman_int = roman_int - 1 * 2;
                if(((s.charAt(i) == 'L') || (s.charAt(i) == 'C')) && (s.charAt(i - 1) == 'X'))
                    roman_int = roman_int - 10 * 2;
                if(((s.charAt(i) == 'D')||(s.charAt(i) == 'M'))&&(s.charAt(i - 1) == 'C'))
                    roman_int = roman_int - 100 * 2;
            }
        }
        return roman_int;
    }
}

// 做法2：
class Solution {
    Map<Character, Integer> symbolValues = new HashMap<Character, Integer>() {{
        put('I', 1);
        put('V', 5);
        put('X', 10);
        put('L', 50);
        put('C', 100);
        put('D', 500);
        put('M', 1000);
    }};

    public int romanToInt(String s) {
        int ans = 0;
        int n = s.length();
        for (int i = 0; i < n; ++i) {
            int value = symbolValues.get(s.charAt(i));
            if (i < n - 1 && value < symbolValues.get(s.charAt(i + 1))) {
                ans -= value;
            } else {
                ans += value;
            }
        }
        return ans;
    }
}