题目如下:
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”.
解题思路:
思路1:第一次使用了一个临时的char二维数组,具体的方法是遍历整个字符串,先将整行都有字符的填满,然后再对有空格的填满,对数据一列一列的操作。
char[][] zig = new char[numRows][s.length()];
String str = "";
int i = 0;
int k = 0;
while(i < s.length()){
for (int j = 0; j < numRows && k < s.length(); j++)
zig[j][i] = s.charAt(k++);
for (int j = numRows - 2; j > 0 && k < s.length(); j--)
zig[j][++i] = s.charAt(k++);
i++;
}
for (int j = 0; j < zig.length; j++) {
for (int j2 = 0; j2 < zig[j].length; j2++) {
if(Character.isLetterOrDigit(zig[j][j2])){
str += zig[j][j2];
}
}
}
return str;
但是,问题来了对于有其他字符的字符串处理起来就明显吃力,比如“,。”等Character.isLetterOrDigit()会将非字母和数字类型的其他字符过滤掉,因此,该方法提交之后Error。
思路2:沿着思路1的方法走,使用一个字符串数组,再赋值的时候就将空格等其他字符去除掉。
提交的代码:
if(numRows == 1) return s;
String[] zig = new String[numRows];
String str = "";
for (int i = 0; i < zig.length; i++) zig[i] = "";
for( int k = 0; k < s.length();){
for (int j = 0; j < numRows && k < s.length(); j++)
zig[j] += s.charAt(k++);
for (int j = numRows - 2; j > 0 && k < s.length(); j--)
zig[j] += s.charAt(k++);
}
for (int i = 0; i < zig.length; i++) str += zig[i];
return str;