[leetcode]ZigZag Conversion

ZigZag Conversion

Difficult：Medium

The string "PAYPALISHIRING" is written in a zigzag pattern on a given number of rows like this: (you may want to display this pattern in a fixed font for better legibility)

P   A   H   NA P L S I I GY   I   R

And then read line by line: "PAHNAPLSIIGYIR"

Write the code that will take a string and make this conversion given a number of rows:

string convert(string text, int nRows);

convert("PAYPALISHIRING", 3) should return "PAHNAPLSIIGYIR".

分析：

问题就是要Z字型输出字符串。

我的实现方法是按要求的行数创建临时的字符串（例如当行数等于3时有s1、s2、s3三个字符串），遍历输入的字符串的时候，按Z字型的顺序把每一位字符放接到临时字符串后边（比如第1个字符接s1后面，第2个接s2，第3个接s3，第4个就又接到s2，第5个接s1，第6个接s2.....），最后再把临时字符串接在一起就好了

 string convert(string s, int numRows) { vector<string> v_result; for (int i = 0; i < numRows; i++){ v_result.push_back("");//按行数生成临时字符串 } int step = 1;//初始步进1，步进为1或-1 if (numRows < 2)//如果行数为1则步进为0 step = 0; int current = 0; for (int i = 0; i < s.size(); i++){ stringstream stream; stream << s.at(i); string tmp = stream.str(); v_result.at(current).append(tmp);//接到相应的临时字符串后面 current = current + step;//加上步进 if (current == 0 || current == numRows - 1)//当current为0或行数时，反转步进 step = step*-1; } string result = ""; for (int i = 0; i < numRows; i++){ result.append(v_result.at(i)); } return result; }