【12】Roman to Integer 罗马数字转化成整数
//第一，如果当前数字是最后一个数字，或者之后的数字比它小的话，则加上当前数字
//第二，其他情况则减去这个数字

class Solution {
public:

int romanToInt(string s) {

int res = 0;

unordered_map m{{'I', 1}, {'V', 5}, {'X', 10}, {'L', 50}, {'C', 100}, {'D', 500}, {'M', 1000}};

for (int i = 0; i < s.size(); ++i) {

int val = m[s[i]];

if (i == s.size() - 1 || m[s[i+1]] <= m[s[i]]) res += val;

else res -= val;

}

return res;

}
};

[6]Zigzag Converesion
将字符串 "PAYPALISHIRING" 以Z字形排列成给定的行数：
P

A

H

N
A P L S I I G
Y

I

R
之后从左往右，逐行读取字符："PAHNAPLSIIGYIR"
实现一个将字符串进行指定行数变换的函数:
string convert(string s, int numRows);
示例 1:
输入: s = "PAYPALISHIRING", numRows = 3
输出: "PAHNAPLSIIGYIR"
示例 2:
输入: s = "PAYPALISHIRING", numRows = 4
输出: "PINALSIGYAHRPI"
解释:
P

I

N
A

L S I G
Y A

H R
P

I

//Java解法
string convert(string s, int numRows)
{

if(numRows <= 1) return s;

string result = "";

int cycle = 2 * numRows - 2;

for(int i = 0; i < numRows; ++i)

{

for(int j = i; j < s.length(); j = j + cycle){

result = result + s[j];

int secondJ = (j - i) + cycle - i;

if(i != 0 && i != numRows-1 && secondJ < s.length())

result = result + s[secondJ];

}

}

return result;
}

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