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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".
思路，用一个idx 从0-->(nRows-1)---->0 来回往返，将对应的字符加入到strs[idx]的字符串中，最后汇总所有的字符串，
注意，当nRows == 1的情况需要特使处理，否则idx-- 后，idx变成-1，会越界crash
class Solution { public: string convert(string s, int nRows) { if(nRows == 1) return s; vector<string> strs; strs.resize(nRows); string result; int idx = 0; bool upOrDown = false;// false--down, true--up for(size_t i = 0; i< s.size(); i++) { strs[idx] += s[i]; if(upOrDown == false) { if(idx == (nRows-1)) { upOrDown = true; idx--; } else { idx++; } } else { if(idx == 0) { upOrDown = false; idx++; } else { idx--; } } } for(size_t i = 0; i< strs.size(); i++) { result += strs[i]; } return result; } };
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原文地址：http://www.cnblogs.com/diegodu/p/4250016.html
