LeetCode算法入门- Roman to Integer && Integer to Roman -day8

LeetCode算法入门- Roman to Integer -day8

Roman to Integer：

1. 题目描述：
   Roman numerals are represented by seven different symbols: I, V, X, L, C, D and M.

Symbol　　　　Value
I　　　　　　　　1
V　　　　　　　　5
X　　　　　　　　10
L　　　　　　　　50
C　　　　　　　　100
D　　　　　　　　500
M　　　　　　　　1000

For example, two is written as II in Roman numeral, just two one’s added together. Twelve is written as, XII, which is simply X + II. The number twenty seven is written as XXVII, which is XX + V + II.

Roman numerals are usually written largest to smallest from left to right. However, the numeral for four is not IIII. Instead, the number four is written as IV. Because the one is before the five we subtract it making four. The same principle applies to the number nine, which is written as IX. There are six instances where subtraction is used:

I can be placed before V (5) and X (10) to make 4 and 9.
X can be placed before L (50) and C (100) to make 40 and 90.
C can be placed before D (500) and M (1000) to make 400 and 900.
Given a roman numeral, convert it to an integer. Input is guaranteed to be within the range from 1 to 3999.

Example 1:

Input: “III”
Output: 3

Example 2:

Input: “IV”
Output: 4

Example 3:

Input: “IX”
Output: 9

Example 4:

Input: “LVIII”
Output: 58
Explanation: L = 50, V = 5, III = 3.

Example 5:

Input: “MCMXCIV”
Output: 1994
Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.

2. 思路分析：
   题目的意思是将罗马数字转成普通int，并且题目给出了罗马数字的表示方式。

思路就是遍历罗马字母，如果当前的罗马字母表示的数字比右边的大，则直接累加即可，如果比右边的小，则需要减去当前罗马字母表示的数字。

代码如下：

class Solution {
    public int romanToInt(String s) {
        int len = s.length();
        //利用HashMap结构来代替if/switch语句
        Map map = new HashMap<>();
        map.put('I',1);
        map.put('V',5);
        map.put('X',10);
        map.put('L',50);
        map.put('C',100);
        map.put('D',500);
        map.put('M',1000);
        int result = 0;
        for(int i = 0; i < len; i++){
            //注意这里要对i进行判断，以免出界
            if(i == len - 1 || map.get(s.charAt(i)) >= map.get(s.charAt(i + 1))){
                result += map.get(s.charAt(i));
            }else{
                //这里也设置的很巧妙，例如LVX = 50 - 5 + 10 = 50 +（10 - 5）
                result -= map.get(s.charAt(i));
            }
        }
        return result;
    }
}

题目衍生：
Integer to Roman

Roman numerals are represented by seven different symbols: I, V, X, L, C, D and M.

Symbol Value
I 1
V 5
X 10
L 50
C 100
D 500
M 1000
For example, two is written as II in Roman numeral, just two one’s added together. Twelve is written as, XII, which is simply X + II. The number twenty seven is written as XXVII, which is XX + V + II.

Roman numerals are usually written largest to smallest from left to right. However, the numeral for four is not IIII. Instead, the number four is written as IV. Because the one is before the five we subtract it making four. The same principle applies to the number nine, which is written as IX. There are six instances where subtraction is used:

I can be placed before V (5) and X (10) to make 4 and 9.
X can be placed before L (50) and C (100) to make 40 and 90.
C can be placed before D (500) and M (1000) to make 400 and 900.
Given an integer, convert it to a roman numeral. Input is guaranteed to be within the range from 1 to 3999.

Example 1:

Input: 3
Output: “III”
Example 2:
Input: 4
Output: “IV”
Example 3:
Input: 9
Output: “IX”
Example 4:
Input: 58
Output: “LVIII”
Explanation: L = 50, V = 5, III = 3.
Example 5:
Input: 1994
Output: “MCMXCIV”
Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.

通过递归来解题：

class Solution {
public String intToRoman(int num) {
        if(num>=1000) return "M"+intToRoman(num-1000);
    //记得添加900，400，90，40，9，4这些范围，不然4它可能表示为IIII，但是我们要的是IV
        if(num>=900) return "CM"+intToRoman(num-900);
        if(num>=500) return "D"+intToRoman(num-500);
        if(num>=400) return "CD"+intToRoman(num-400);
        if(num>=100) return "C"+intToRoman(num-100);
        if(num>=90) return "XC"+intToRoman(num-90);
        if(num>=50) return "L"+intToRoman(num-50);
        if(num>=40) return "XL"+intToRoman(num-40);
        if(num>=10) return "X"+intToRoman(num-10);
        if(num>=9) return "IX"+intToRoman(num-9);
        if(num>=5) return "V"+intToRoman(num-5);
        if(num>=4) return "IV"+intToRoman(num-4);
        if(num>=1) return "I"+intToRoman(num-1);
        return "";
    }
}

非递归亦可以实现：

class Solution {
    //非递归解法
public String intToRoman(int num) {
    //利用两个数组来存储罗马数字和其值，通过下标来进行对应
    String result = "";
    //同样注意的是要扩展罗马数字的范围，不然4会被表示为IIII，而不是IV
    String[] str = {"M","CM","D","CD","C","XC","L","XL","X","IX","V","IV","I"};
    int[] val = {1000,900,500,400,100,90,50,40,10,9,5,4,1};
    for(int i = 0; i < 13; i++){
        while(num >= val[i]){
            num = num - val[i];
            result =result + str[i];
        }
    }
    return result;
    }
}