【LeetCode算法-13】Roman to Integer
LeetCode第13题
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.
废话不多说,直接上代码,姿势一定要好看
class Solution {
private int[] num;
public int romanToInt(String s) {
//先将罗马字母转为数字
stringToNum(s);
//检查左小右大的情况
checkLeftDigit();
//求和
int sum = 0;
for(int l = 0;l<num.length;l++){
sum+=num[l];
}
return sum;
} public void stringToNum(String s){
num = new int[s.length()];
for(int i=0;i<s.length();i++){
switch(s.charAt(i)){
case 'I':
num[i]=1;
break;
case 'V':
num[i]=5;
break;
case 'X':
num[i]=10;
break;
case 'L':
num[i]=50;
break;
case 'C':
num[i]=100;
break;
case 'D':
num[i]=500;
break;
case 'M':
num[i]=1000;
break;
default:
num[i]=0;
break;
}
}
} public void checkLeftDigit(){
for(int j = 0;j<num.length;j++){
for(int k = j+1;k<num.length;k++){
if(num[j]==1 &&(num[k]==5 || num[k]==10)){
num[j] *= (-1);
}else if(num[j]==10 &&(num[k]==50 || num[k]==100)){
num[j] *= (-1);
}else if(num[j]==100 &&(num[k]==500 || num[k]==1000)){
num[j] *= (-1);
}
}
}
}
}
1.因为是手写,api会写错,String的长度是length()不是length;数组的长度是length不是size()
2.一开始我考虑会不会有IXC(109?91?89?)的情况,后来审题发现罗马字母正常情况左大右小,何况109是CIX,91是XCI,89是LXXXIX;所以不会有IXC的情况,就不用判断了
3.其实Char也能通过ASCII来判断大小和差值,不过最后还是要转成INT相加,所以我就先转成INT了
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