Given a roman numeral, convert it to an integer.
Input is guaranteed to be within the range from 1 to 3999.
» Solve this problem[解题思路]
从前往后扫描,用一个临时变量记录分段数字。
- 如果当前比前一个大,说明这一段的值应该是当前这个值减去上一个值。比如IV = 5 – 1
- 否则,将当前值加入到结果中,然后开始下一段记录。比如VI = 5 + 1, II=1+1
[Code]
1: inline int c2n(char c) {
2: switch(c) {
3: case 'I': return 1;
4: case 'V': return 5;
5: case 'X': return 10;
6: case 'L': return 50;
7: case 'C': return 100;
8: case 'D': return 500;
9: case 'M': return 1000;
10: default: return 0;
11: }
12: }
13: int romanToInt(string s) {
14: // Start typing your C/C++ solution below
15: // DO NOT write int main() function
16: int result=0;
17: for(int i =0; i< s.size(); i++)
18: {
19: if(i>0&& c2n(s[i]) > c2n(s[i-1]))
20: {
21: result +=(c2n(s[i]) - 2*c2n(s[i-1]));
22: }
23: else
24: {
25: result += c2n(s[i]);
26: }
27: }
28: return result;
29: }
关联题:
http://fisherlei.blogspot.com/2012/12/leetcode-integer-to-roman.html
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