一天一道LeetCode系列
(一)题目
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”.
题意:
(二)解题
已知输入条件:初始string s和numRows。可以得出循环间隔gap = 2*numRows-2。
统计规律:
- 第0行 从j=0开始,间距gap,取s[j]
- 第1~numRows-2行 从j=1开始,间距gap=(j%gap)<numRows?(gap-2(j%gap)):(2*gap-2(j%gap))
,取s[j]
- 第numRows-1行,间距gap,取s[j]
代码如下:
class Solution {
public:
string convert(string s, int numRows) {
string result;
int gap=2*numRows-2; //初始化gap
if(numRows == 1) return s;//rows为1,直接返回
for(int i = 0 ; i < numRows ; i++)
{
int j = i ;
bool flag = true;
if(i == 0 || i == numRows-1)//如果第0行或最后一行
{
int j = i;
while(j<s.length())
{
result+=s[j];
j += gap;//间距为gap
}
}
else{
int j = i;
while(j<s.length()){
result+=s[j];
j += (j%gap)<numRows?(gap-2*(j%gap)):(2*gap-2*(j%gap));
}
}
}
return result;
}
};