字符串|【LeetCode】ZigZag Conversion

ZigZag Conversion
Total Accepted: 5118 Total Submissions: 22667 My Submissions
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)
PAHN
A P L S I I G
YIR
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".
这个其实是个纯数学题,找规律。
我们来看几组数据
nRows = 1,结构是这样的
字符串|【LeetCode】ZigZag Conversion
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nRows = 2,结构是这样的
字符串|【LeetCode】ZigZag Conversion
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【字符串|【LeetCode】ZigZag Conversion】nRows = 3,结构是这样的
字符串|【LeetCode】ZigZag Conversion
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nRows = 4,结构是这样的
字符串|【LeetCode】ZigZag Conversion
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nRows = 5,结构是这样的
字符串|【LeetCode】ZigZag Conversion
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根据上述图表,有这样的规律。
针对每行来说,包含的字符串间隔是2*nRows-2;
当nRows = 0和nRows = len-1时,不需要处理中间的字符。
其余行,则针对每列来说,还需要增加当前列+2*nRows-4,一直到2为止。
Java AC

public class Solution { public String convert(String s, int nRows) { if(nRows == 1){ return s; } int len = s.length(); StringBuffer sb = new StringBuffer(); char array[] = s.toCharArray(); int rowLen = 2 * nRows - 2; int colLen = rowLen + 2; for(int i = 0; i < nRows; i++){ int j = i; colLen -= 2; while(j < len){ if(i == 0 || i == nRows - 1){ sb.append(String.valueOf(array[j])); j += rowLen; }else{ sb.append(String.valueOf(array[j])); int tempLen = j + colLen; if(tempLen < len){ sb.append(String.valueOf(array[tempLen])); } j += rowLen; } } } return sb.toString(); } }


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