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ZigZag Conversion

2024-07-21 13:28:46   219人阅读

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".

思路：若为第一行或最后一行（即i=0 or i=nRow-1），则同一行的相邻元素索引相差2*(nRow-1)；若为中间行(即0<i<nRow-1)，则相邻索引相差为2*(nRow-1-i)和2i的交替。

 1 class Solution { 2 public: 3     string convert( string s, int nRows ) { 4         if( --nRows <= 0 ) { return s; } 5         int slen = s.length(); 6         string zigzag = ""; 7         for( int i = 0; i <= nRows; ++i ) { 8             if( i == 0 || i == nRows ) { 9                 for( int k = i; k < slen; k += 2*nRows ) {10                     zigzag += s[k];11                 }12             } else {13                 for( int k = i; k < slen; k += 2*nRows ) {14                     zigzag += s[k];15                     if( k+2*(nRows-i) < slen ) { zigzag += s[k+2*(nRows-i)]; }16                 }17             }18         }19         return zigzag;20     }21 };

 

ZigZag Conversion


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