[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

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 integerdef romanToInt(self, s):val = {'I': 1, 'V': 5, 'X': 10, 'L': 50, 'C': 100, 'D': 500, 'M': 1000}ret = 0for 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

不要因为世态变迁而埋怨,不要因为命运多舛而怨恨.