Roman to Integer
Given a roman numeral, convert it to an integer.
Input is guaranteed to be within the range from 1 to 3999.
SOLUTION 1:
思路：
从后往前遍历罗马数字，如果某个数比前一个数小，则把该数在结果中减掉；
反之，则在结果中加上当前这个数；
GITHUB 代码：
https://github.com/yuzhangcmu/LeetCode_algorithm/blob/master/string/RomanToInt.java
描述： Given an input string (`s`) and a pattern (`p`), implement regular expression match
Topic String Dynamic Programming Backtracking Description [https://leetc
public class No10 { public static void main(String[] args) {
Implement regular expression matching with support for `'.'` and `''`. '.' Matches
这个题目比较常见，但是难度还是比较大的。我们先来看看brute force怎么解决。基本思路就是先看字符串s和p的从i和j开始的子串是否匹配，用递归的方法直到串的最后
Implement regular expression matching with support for `'.'` and `''`. '.' Matches
Implement regular expression matching with support for `'.'` and `''`. '.' Matches
Implement regular expression matching with support for ‘.’ and ‘\’. ‘.’ Matches any sin
Roman to Integer Given a roman numeral, convert it to an integer. Input is guaran
深藏阁楼爱情的钟
た 入场券
我就是我
还没有评论，来说两句吧...