[Leetcode] 013. Roman to Integer (easy) (C++/java/python)

Index: [Leetcode] leetcode key index (C++/JAVA/PYTHON/SQL)
Github:https://github.com/illuz/leetcode

013.roman_to_integer (Easy) links：

Title: https://oj.leetcode.com/problems/roman-to-integer/
Code (GitHub): Https://github.com/illuz/leetcode

Test Instructions：

Convert the Roman number to decimal.

Analysis：

With 012. Integer to Roman (Medium), just know the conversion rules.

Code：

C++:

Class Solution {Private:int val[255];void init () {val[' I '] = 1; val[' V '] = 5; val[' X '] = ten; val[' L '] = 50;val[' C '] = 100; val[' D '] = 500; val[' M '] = 1000;} Public:    int Romantoint (string s) {init (); int ret = 0;for (int i = 0; i < s.size (); i++) {if (i > 0 && VA L[s[i]] > Val[s[i-1]) {ret + val[s[i]]-2 * val[s[i-1]];} else {ret + = Val[s[i]];}} return ret;    }};

Java:

public class Solution {    private int[] val = new int[255];    private void Init () {        val[' I '] = 1; val[' V '] = 5; val[' X '] = ten; val[' L '] =        ; val[' C '] = 100; val[' D '] = 500; val[' M '] = +;    }    public int Romantoint (String s) {        init ();        int ret = 0;        for (int i = 0; i < s.length (); i++) {            if (i > 0 && val[s.charat (i)] > Val[s.charat (i-1)]) {                R ET + = Val[s.charat (i)]-2 * Val[s.charat (I-1);            } else {                ret + = Val[s.charat (i)];}        }        return ret;    }}

Python:

Class solution:    # @return An integer    def romantoint (self, s):        val = {' I ': 1, ' V ': 5, ' X ': Ten, ' L ':-10 0, ' D ': +, ' M ': +}        ret = 0 for        i in range (len (s)):            If i > 0 and val[s[i]] > val[s[i-1]:                ret + = Val[s[i]]-2 * val[s[i-1] [            else:                ret + val[s[i]]        return RET

[Leetcode] 013. Roman to Integer (easy) (C++/java/python)