题目描述 罗马数字包含以下七种字符:I, V, X, L,C,D 和 M. 字符 数值 I 1 V 5 X 10 L 50 C 100 D 500 M 1000 例如, 罗马数字 2 写做 II ,即为两个并列的 1.12 写做 XII ,即为 X + II . 27 写做 XXVII, 即为 XX + V + II . 通常情况下,罗马数字中小的数字在大的数字的右边.但也存在特例,例如 4 不写做 IIII,而是 IV.数字 1 在数字 5 的左边,所表示的数等于大数 5 减小数 1 得到的数…
13. Roman to Integer Easy Roman numerals are represented by seven different symbols: I, V, X, L, C, D and M. Symbol Value I 1 V 5 X 10 L 50 C 100 D 500 M 1000 For example, two is written as II in Roman numeral, just two one's added together. Twelve i…
题意: Roman numerals are represented by seven different symbols: I, V, X, L, C, D and M. Symbol Value I 1 V 5 X 10 L 50 C 100 D 500 M 1000 For example, two is written as II in Roman numeral, just two one's added together. Twelve is written as, XII, whi…
Roman numerals are represented by seven different symbols: I, V, X, L, C, D and M. Symbol Value I 1 V 5 X 10 L 50 C 100 D 500 M 1000 For example, two is written as II in Roman numeral, just two one's added together. Twelve is written as, XII, which i…
1.题目: 原题:Given a roman numeral, convert it to an integer. Input is guaranteed to be within the range from 1 to 3999. Subscribe to see which companies asked this questio 解析:给出一个罗马数字,要求把其转换为一个整数.输入范围在1到3999内. 罗马数字的规则如下: 罗马数字 I V X L C D M 代表的阿拉伯数字 1 5…
问题描述: Given a roman numeral, convert it to an integer. Input is guaranteed to be within the range from 1 to 3999. 解题思路: 上一题反过来就可以了,其实觉得这两道题很没有意思. 代码如下: public class Solution { public int romanToInt(String s) { int result; if (s == null || s.length()…
题目要求 Roman numerals are represented by seven different symbols: I, V, X, L, C, Dand M. Symbol Value I 1 V 5 X 10 L 50 C 100 D 500 M 1000 For example, two is written as II in Roman numeral, just two one's added together. Twelve is written as, XII, whi…
Roman numerals are represented by seven different symbols: I, V, X, L, C, D and M. Symbol Value I 1 V 5 X 10 L 50 C 100 D 500 M 1000 For example, two is written as II in Roman numeral, just two one's added together. Twelve is written as, XII, which i…
Given a roman numeral, convert it to an integer. Input is guaranteed to be within the range from 1 to 3999. 解法: 只要考虑两种情况即可: 第一,如果当前数字是最后一个数字,或者之后的数字比它小的话,则加上当前数字 第二,其他情况(即为4或者9,这种情况下才可能出现后面的罗马数字比前面的大),则减去这个数字 public class Solution { public int romanT…
如果某一个字母代表的数字大于上一个字母代表的数字小,那么加上这个数字,否则,减去两倍的前一个数字,然后加上这一位数字. public class Solution { private static char[][] chars = {{'I', 'V'}, {'X', 'L'}, {'C', 'D'}, {'M', 'O'}}; private static HashMap<Character, Integer> map = null; private static void init() {…