标签：
leetcode - ZigZag Conversion
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 N A P L S I I G Y 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".
1 class Solution { 2 public: 3 string convert(string s, int numRows) { 4 vector<string> result(numRows); 5 bool increase = true; 6 string output; 7 string::iterator iter = s.begin(); 8 int i = 0; 9 if(s.size()<numRows || numRows == 1){ //special cases 10 return s; 11 } 12 result[0].push_back(*iter);// the first character 13 iter++; 14 while(iter != s.end()){ 15 if(increase){ 16 i++; 17 } 18 else{ 19 i--; 20 } 21 if(i==(numRows-1) || i==0){ //here, avoid that i=0 for the first charater, it will become false.(it should be ture) 22 increase = !increase; 23 } 24 result[i].push_back(*iter); 25 iter++; 26 } 27 for(int j=0; j<numRows; j++){ 28 output += result[j]; 29 } 30 return output; 31 } 32 };
思路： 建立numRows维vector存放每行数据，然后开始遍历，i控制行号，行号从0-（numRow-1），再到0，一直循环。
直到全部字符遍历完，按行输出即可。
要注意特殊情况，行数大于字符总数，或者行数等于一（好像已经考虑过了。。）。
标签：
原文地址：http://www.cnblogs.com/shnj/p/4572025.html
