Leetcode Roman to Integer (C,c++,java,python)

Problem:

Given a Roman numeral, convert it to an integer.

Input is guaranteed to being within the range from 1 to 3999.

Solution: Time complexity O (n) topic: In contrast to 12, give a Roman numeral, ask to convert to decimal number problem solving idea:
Java source code (spents 749ms):

public class Solution {public    int romantoint (String s) {        int index=0,num=0,temp=0;        while (Index<s.length ()) {            char c=s.charat (index++);            Switch (c) {case                ' I ': num+=1;temp=1;break;                Case ' V ': num+=temp==1?3:5;break;                Case ' X ': num+=temp==1?8:10;temp=10;break;                Case ' L ': num+=temp==10?30:50;break;                Case ' C ': num+=temp==10?80:100;temp=100;break;                Case ' D ': num+=temp==100?300:500;break;                Case ' M ': num+=temp==100?800:1000;break;            }        }        return num;    }}

C Language Source code (spents 18ms):

int Romantoint (char* s) {    int num=0,temp=0;    while (*s) {        switch (*s) {case            ' I ': num+=1;temp=1;break;            Case ' V ': num+=temp==1?3:5;break;            Case ' X ': num+=temp==1?8:10;temp=10;break;            Case ' L ': num+=temp==10?30:50;break;            Case ' C ': num+=temp==10?80:100;temp=100;break;            Case ' D ': num+=temp==100?300:500;break;            Case ' M ': num+=temp==100?800:1000;break;        }        s++;    }    return num;}

C + + source code (spents 58ms):

Class Solution {public:    int Romantoint (string s) {        int index=0,num=0,temp=0;        while (Index<s.size ()) {            char c=s[index++];            Switch (c) {case                ' I ': num+=1;temp=1;break;                Case ' V ': num+=temp==1?3:5;break;                Case ' X ': num+=temp==1?8:10;temp=10;break;                Case ' L ': num+=temp==10?30:50;break;                Case ' C ': num+=temp==10?80:100;temp=100;break;                Case ' D ': num+=temp==100?300:500;break;                Case ' M ': num+=temp==100?800:1000;break;            }        }        return num;    }};

Python source code (spents 138ms):

Class solution:    # @param {string} s    # @return {integer}    def romantoint (self, s):        index=0;num=0;temp=0< C4/>while Index<len (s):            c = s[index];index+=1            if c== ' I ': num+=1;temp=1            elif c== ' V ': num+=3 if temp==1 Else 5            elif c== ' X ': num+=8 if temp==1 else 10;temp=10            elif c== ' L ': num+=30 if temp==10 else            elif c== ' C ': num+ =80 if temp==10 else 100;temp=100            elif c== ' D ': num+=300 if temp==100 else            elif c== ' M ': num+=800 if temp==100 E LSE        return NUM

Leetcode Roman to Integer (C,c++,java,python)