13. Roman to Integer //罗马数字转整数
- Total Accepted: 121305
- Total Submissions: 280158
- Difficulty: Easy
- Contributors: Admin
Given a roman numeral, convert it to an integer.
Input is guaranteed to be within the range from 1 to 3999.
---------------------------------第一次觉得自己这次代码很简洁-------------------------------
时间复杂度O(n)，参照12题（见下）逆向的Top Solution，得出规律——只要处理好CD，CM，XL，XC，IV，IX这几个之后其他的按单个的累加即可。
//String M[] = {"", "M", "MM", "MMM"};
//String C[] = {"", "C", "CC", "CCC", "CD", "D", "DC", "DCC", "DCCC", "CM"};
//String X[] = {"", "X", "XX", "XXX", "XL", "L", "LX", "LXX", "LXXX", "XC"};
//String I[] = {"", "I", "II", "III", "IV", "V", "VI", "VII", "VIII", "IX"};
if(i+1i+1
//本来以为去掉会提速的，结果并没有
class Solution {
public:
int romanToInt(string s) {
int result=0;
for(int i=0;
i运行结果：
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难得一次击败了80%+的人，mark一下
祝刷题愉快~
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【LeetCode|【LeetCode013算法/编程练习C++】Roman to Integer //难得击败了80%的人】
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