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

索引：[LeetCode] Leetcode 题解索引 (C++/Java/Python/Sql)
Github: https://github.com/illuz/leetcode

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013.Roman_to_Integer (Easy)

链接：

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

题意：

把罗马数转为十进制。

分析：

跟 012. Integer to Roman (Medium) 一样，只要知道转化规则就行了。

代码：

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