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算法的重要性，我就不多說(shuō)了吧，想去大廠，就必須要經(jīng)過(guò)基礎(chǔ)知識(shí)和業(yè)務(wù)邏輯面試+算法面試。所以，為了提高大家的算法能力，這個(gè)公眾號(hào)后續(xù)每天帶大家做一道算法題，題目就從LeetCode上面選！

今天和大家聊的問(wèn)題叫做羅馬數(shù)字轉(zhuǎn)整數(shù)?，我們先來(lái)看題面：

https://leetcode-cn.com/problems/roman-to-integer/

題意

樣例

示例?1:

輸入:?"III"
輸出:?3
示例?2:

輸入:?"IV"
輸出:?4
示例?3:

輸入:?"IX"
輸出:?9
示例?4:

輸入:?"LVIII"
輸出:?58
解釋: L =?50, V=?5, III =?3.

題解

哈希解法：用map建立起來(lái)鍵值對(duì),每個(gè)可能組成的羅馬數(shù)字一一對(duì)應(yīng)為整數(shù),
利用子字符串方法subStrirng拿出兩位字符串,然后在map中查找是否有這個(gè)key,如果有則拿出key鍵所對(duì)應(yīng)的value值,用一個(gè)變量記錄為整數(shù)的值加這個(gè)value值,如果存在兩位的羅馬數(shù)字,則角標(biāo)加2,如果不存在兩位羅馬數(shù)字,即是加1.

public?static?int?romanToInt(String s)?{
????????HashMap map?= new?HashMap<>();
????????int?result = 0;
????????map.put("I", 1);
????????map.put("IV", 4);
????????map.put("V", 5);
????????map.put("IX", 9);
????????map.put("X", 10);
????????map.put("XL", 40);
????????map.put("L", 50);
????????map.put("XC", 90);
????????map.put("C", 100);
????????map.put("CD", 400);
????????map.put("D", 500);
????????map.put("CM", 900);
????????map.put("M", 1000);

????????for?(int?i = 0; i < s.length(); ) {
????????????if?(i + 1?< s.length() && map.containsKey(s.substring(i, i + 2))) {
????????????????result += map.get(s.substring(i, i + 2));
????????????????i += 2;
????????????} else?{
????????????????result += map.get(""?+ s.charAt(i));
????????????????i++;
????????????}
????????}
????????return?result;
????}