Roman to Integer(罗马数转为整数)leetcode13
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.
code1:
public int romanToInt(String s) {
/*int[] nums={1000,900,500,400,100,90,50,40,10,9,5,4,1};
String[] romans={"M","CM","D","CD","C","XC","L","XL","X","IX","V","IV","I"};
int count=0;
int result=0;
while(count<s.length()){
String res1="";
if(count<s.length()-1)
res1=s.substring(count,count+2);
String res2=s.substring(count,count+1);
for(int i=0;i<romans.length;i++){
if(romans[i].equals(res1)){
count+=2;
result+=nums[i];
break;
}else if(romans[i].equals(res2)){
count++;
result+=nums[i];
break;
}
}
}
return result;
}
code2:
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;//从前往后遍历或者下面那种从后往前遍历
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];*/
int sum=nums[nums.length-1];
for(int i=nums.length-2;i>=0;i--){
if(nums[i]<nums[i+1])
sum-=nums[i];
else
sum+=nums[i];
}
return sum;
}
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