[LeetCode] 013. Roman to Integer (Easy) (C ++/Java/Python), leetcodepython

[LeetCode] 013. Roman to Integer (Easy) (C ++/Java/Python), leetcodepython

Index: [LeetCode] Leetcode index (C ++/Java/Python/SQL)
Github: https://github.com/illuz/leetcode

013. Roman_to_Integer (Easy) Link:

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

Question:

Convert the number of Rome to decimal.

Analysis:

Like 012. Integer to Roman (Medium), you only need to know the conversion rules.

Code:

C ++:

class Solution {private:int val[255];void init() {val['I'] = 1; val['V'] = 5; val['X'] = 10; 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 && val[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'] = 10; 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.length(); i++) {            if (i > 0 && val[s.charAt(i)] > val[s.charAt(i - 1)]) {                ret += 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': 10, 'L': 50, 'C': 100, 'D': 500, 'M': 1000}        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