Roman to Integer - LeetCode

题目链接
注意点
解法
小结

题目链接

注意点

大的数字在小的数字左边是正常情况，要考虑的是小的数字在大的数字右边的那8种情况

解法

解法一：从右至左，对于出现的每个字符逐个判断，累加。时间复杂度为O(n)

class Solution {
public:
    int romanToInt(string s) {
        int i,n = s.size(),anw = 0;
        for(i = n-1;i >= 0;i--)
        {
            if(s[i] == 'M')
            {
                if(s[i-1] == 'C')
                {
                    anw += 900;
                    i--;
                }
                else
                {
                    anw += 1000;
                }
            }
            else if(s[i] == 'D')
            {
                if(s[i-1] == 'C')
                {
                    anw += 400;
                    i--;
                }
                else
                {
                    anw += 500;
                }
            }
            else if(s[i] == 'C')
            {
                if(s[i-1] == 'X')
                {
                    anw += 90;
                    i--;
                }
                else
                {
                    anw += 100;
                }
            }
            else if(s[i] == 'L')
            {
                if(s[i-1] == 'X')
                {
                    anw += 40;
                    i--;
                }
                else
                {
                    anw += 50;
                }
            }
            else if(s[i] == 'X')
            {
                if(s[i-1] == 'I')
                {
                    anw += 9;
                    i--;
                }
                else
                {
                    anw += 10;
                }
            }
            else if(s[i] == 'V')
            {
                if(s[i-1] == 'I')
                {
                    anw += 4;
                    i--;
                }
                else
                {
                    anw += 5;
                }
            }
            else if(s[i] == 'I')
            {
                anw += 1;
            }
            //printf("%d\n",anw);
        }
        return anw;
    }
};

解法二：从左至右。如果之前的数字小于当前的就减去之前的数字，否则就加上。最后一个字符无论如何都加上。时间复杂度O(n)

class Solution {
public:
    int romanToInt(string s) {
        map<char,int> m;
        m['I'] = 1;
        m['V'] = 5;
        m['X'] = 10;
        m['L'] = 50;
        m['C'] = 100;
        m['D'] = 500;
        m['M'] = 1000;
        int i,n = s.size(),anw = 0;
        int before = m[s[0]],now = 0;
        for(i = 1;i < n;i++)
        {
            now = m[s[i]];
            if(before < now)
            {
                anw -= before;
            }
            else
            {
                anw += before;
            }
            before = now;
        }
        anw += before;
        return anw;
    }
};