如题:罗马数字到整形数据的转换
input:III
output:3
input:IV
output:4
input:IX
output:9
input:LVIII
output:58
input:MCMXCIV
output:1994
int romanToInt(string s) {
int num = 0;
vector<int> vec;
for (int i = 0; i < s.size(); ++i) {
if (s[i] == 'I') {
vec.push_back(1);
continue;
}
if (s[i] == 'V') {
vec.push_back(5);
continue;
}
if (s[i] == 'X') {
vec.push_back(10);
continue;
}
if (s[i] == 'L') {
vec.push_back(50);
continue;
}
if (s[i] == 'C') {
vec.push_back(100);
continue;
}
if (s[i] == 'D') {
vec.push_back(500);
continue;
}
if (s[i] == 'M') {
vec.push_back(1000);
continue;
}
}
for (int i = 0; i < s.size() - 1; ++i) {
int j = i + 1;
if (vec[i] < vec[j] && vec[j] / vec[i] <= 10) {
vec[i] = vec[j] - vec[i];
vec[j] = 0;
i++;
}
}
for (auto i : vec)
num += i;
return num;
}
第一个for循环将所有罗马数字对应的整型值输入到vector中,第二个循环判断有没有如(IV)这种表示(V-I)的数,如果有则替换掉,最后一个范围for循环将容器中所有数加起来return,复杂度为O(N)