LeetCode算法入門- Roman to Integer -day8

Roman to Integer：

題目描述：
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.

思路分析：
題目的意思是將羅馬數字轉成普通int，并且題目給出了羅馬數字的表示方式。

思路就是遍歷羅馬字母，如果當前的羅馬字母表示的數字比右邊的大，則直接累加即可，如果比右邊的小，則需要減去當前羅馬字母表示的數字。

代碼如下：

class Solution {public int romanToInt(String s) {int len = s.length();//利用HashMap結構來代替if/switch語句Map<Character,Integer> 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++){//注意這里要對i進行判斷，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{//這里也設置的很巧妙，例如LVX = 50 - 5 + 10 = 50 +（10 - 5）result -= map.get(s.charAt(i));}}return result;} }

題目衍生：
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.

通過遞歸來解題：

class Solution { public String intToRoman(int num) {if(num>=1000) return "M"+intToRoman(num-1000);//記得添加900，400，90，40，9，4這些范圍，不然4它可能表示為IIII，但是我們要的是IVif(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 "";} }

非遞歸亦可以實現：

class Solution {//非遞歸解法 public String intToRoman(int num) {//利用兩個數組來存儲羅馬數字和其值，通過下標來進行對應String result = "";//同樣注意的是要擴展羅馬數字的范圍，不然4會被表示為IIII，而不是IVString[] 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;} }

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