Leetcode 13 Roman to Integer
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 beforeV
(5) andX
(10) to make 4 and 9.X
can be placed beforeL
(50) andC
(100) to make 40 and 90.C
can be placed beforeD
(500) andM
(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: C = 100, L = 50, XXX = 30 and III = 3.
Example 5:
Input: "MCMXCIV" Output: 1994 Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.
在做的過程中map的key使用了string,結果一直報錯,之後改為character之後問題解決。
1)
class Solution {
public int romanToInt(String s) {
int sum = 0;
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);
sum = map.get(s.charAt(s.length() - 1));
for(int i = s.length() - 2;i >= 0;i--){
if(map.get(s.charAt(i)) < map.get(s.charAt(i+1))){
sum -= map.get(s.charAt(i));
}else{
sum += map.get(s.charAt(i));
}
}
return sum;
}
}
2)
class Solution {
public int romanToInt(String s) {
char[] chs=s.toCharArray();
int temp=0,sum=0,pre=0;
pre=500000;
for(int i=0;i<chs.length;i++){
char ch=chs[i];
switch(ch){
case 'I' :temp=1;break;
case 'V' :temp=5;break;
case 'X' :temp=10;break;
case 'L' :temp=50;break;
case 'C' :temp=100;break;
case 'D' :temp=500;break;
case 'M' :temp=1000;break;
default:temp=0;break;
}
if(pre<temp)
sum=sum+temp-2*pre;
else
sum+=temp;
pre=temp;
}
return sum;
}
}
相關文章
- LeetCode 13. Roman to Integer C語言LeetCodeC語言
- 13. Roman to Integer
- Leetcode 12 Integer to RomanLeetCode
- Leetcode 12. Integer to RomanLeetCode
- LeetCode Integer to Roman(012)解法總結LeetCode
- LeetCode Roman to Integer(013)解法總結LeetCode
- LeetCode - 解題筆記 - 12 - Integer to RomanLeetCode筆記
- Roman to Integer 羅馬數字轉整數
- Leetcode 7 Reverse IntegerLeetCode
- Leetcode 273 Integer to English WordsLeetCode
- Leetcode 8 String to Integer (atoi)LeetCode
- LeetCode Reverse Integer(007)解法總結LeetCode
- [leetcode] 1394. Find Lucky Integer in an ArrayLeetCode
- LeetCode String to Integer (atoi)(008)解法總結LeetCode
- LeetCode - 解題筆記 - 7 - Reverse IntegerLeetCode筆記
- Leetcode-13LeetCode
- Leetcode 8. String to Integer (atoi) 字串轉整數 (atoi)LeetCode字串
- LeetCode刷題記13-27. 移除元素LeetCode
- LeetCode 13[羅馬數字轉整數]LeetCode
- Integer比較
- LeetCode_Python(13)_羅馬數字轉整數LeetCodePython
- leetcode13題——羅馬數字轉整數LeetCode
- Q13 LeetCode76 最小覆蓋子串LeetCode
- python-leetcode13羅馬數字轉整數PythonLeetCode
- Integer包裝類
- [Java基礎]IntegerJava
- Integer的比較
- 程式碼審查清單:Java併發 - Roman LeventovJava
- 走進 JDK 之 IntegerJDK
- Integer128==128?falseFalse
- ibatis中integer型別BAT型別
- Java Integer的快取策略Java快取
- int和Integer的區別
- Integer 自動拆箱封箱
- Integer轉int出現NullPointExceptionNullException
- int與Integer的區別
- 產品路線圖要避免的10個錯誤 - roman
- JDK原始碼閱讀-Integer類JDK原始碼