共计 741 个字符,预计需要花费 2 分钟才能阅读完成。
Easy 013 Roman to Integer
Description:
将罗马字母的字符串转换为代表的整数
Roman numerals are usually written largest to smallest from left to right. However, there are six instances where subtraction is used:
I can be placed before V (5) and X (10) to make 4 and 9.
X can be placed before L (50) and C (100) to make 40 and 90.
C can be placed before D (500) and M (1000) to make 400 and 900.
Symbol Value
I 1
V 5
X 10
L 50
C 100
D 500
M 1000
My Solution:
-
这题不难,用一个 HashMap 存罗马数字和具体数字的对应关系,然后遍历前后两两比较,该加加,该减减
- 时间复杂度 O(n)
Fast Solution:
public static int romanToInt(String s) {
int num = 0;
int n = s.length();
for (int i = 0; i < n-1; i++) {int curr = map(s.charAt(i)); // 这里 map 是自己写的一个方法,里面用一个 switch,相当于 HashMap 存对应
int next = map(s.charAt(i+1));
num = curr < next ? num - curr : num + curr;
// 当时一直想着用一个 temp 来存减的值,所以没法用 for 就用了 while&point 指针,但其实就是很简单的比后面小的话在总值里面减去就可以,不需要 temp
}
num += map(s.charAt(n-1));
return num;
}
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