LeetCode第[13]题(Java):Roman to Integer
题目:罗马数字转换
题目难度:easy
题目内容: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.
翻译:
例如,2在罗马数字中被写成II,就是两个加在一起。12是写成,XII,也就是X+II。数字二十七是二十七,也就是XX+V+II。
罗马数字通常从左到右写得最大。然而,4的数字不是IIII。相反,数字4被写成IV,因为1在5之前减去它等于4。同样的原则也适用于9号,它被写成IX。有六种情况下使用减法:
I可以在V(5)和X(10)之前放置分别是4和9。
X可以放在L(50)和C(100)之前,分别是40和90。
C可以放置在D(500)和M(1000)之前,分别是400和900。
给定一个罗马数字,把它转换成整数。输入在从1到3999的范围内。
思路:
加法直接加,减法只匹配前后两个字符,所以当cur字符小于下一个字符的时候,使用减法。
用一个Map将各个字符与对应值都存进去,即可进行比较和取值
MyCode:
public int romanToInt(String s) {
if (s == null || s.isEmpty()) {
return -1;
} char[] sChar = s.toCharArray();
Map<Character, Integer> map = new HashMap<Character, Integer>();
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 sum=0;
for(int i=0;i<sChar.length-1;i++){
if(map.get(sChar[i]) < map.get(sChar[i+1]))
sum-=map.get(sChar[i]); // 当前字符比下一个小,则总和减去此数字
else
sum+=map.get(sChar[i]); // 否则直接加上
}
return sum+map.get(sChar[sChar.length-1]); // 因为最后一个没判断,并且无后续,所以必然是加
}
结果:Accept
参考答案:
public int romanToInt(String s) {
int nums[]=new int[s.length()];
for(int i=0;i<s.length();i++){
switch (s.charAt(i)){
case 'M':
nums[i]=1000;
break;
case 'D':
nums[i]=500;
break;
case 'C':
nums[i]=100;
break;
case 'L':
nums[i]=50;
break;
case 'X' :
nums[i]=10;
break;
case 'V':
nums[i]=5;
break;
case 'I':
nums[i]=1;
break;
}
}
int sum=0;
for(int i=0;i<nums.length-1;i++){
if(nums[i]<nums[i+1])
sum-=nums[i];
else
sum+=nums[i];
}
return sum+nums[nums.length-1];
}
答案思路:此处直接将原来的字符数组(字符串)转换成相对应的int数组,这样就不需要一个map了,相对来说这种方法的访问更加简便,但是增加了一轮算法复杂度,不过map的访问要比直接数组访问多一个步骤,所以两者复杂度算下来可能差不多都是O(2n)的样子,而答案这种操作更加简单一些,可以借鉴。
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