LeetCode | 6. ZigZag Conversion(之字形变换)

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:两种.最初想到的是开一个数组,每次像蛇形矩阵一样对数组赋值,最后对每一行每一列遍历,提取出不是空的字符.125ms,几乎是所有题叫中所有提交中最慢的之一了.

第二种,找规律,直接按0~r行,空间复杂度为O(1).29ms(参考LeetCode代码)

虽然这两个算法的复杂度都是O(n),时间差别还挺大.

/*n=numRows

Δ=2n-2 1 2n-1 4n-3

Δ= 2 2n-2 2n 4n-4 4n-2

Δ= 3 2n-3 2n+1 4n-5 .

Δ= . . . . .

Δ= . n+2 . 3n .

Δ= n-1 n+1 3n-3 3n-1 5n-5

Δ=2n-2 n 3n-2 5n-4

*/

//125ms
class Solution {
public:
string convert(string s, int r)
{
int len = s.length();
string res = "";
if (r == 1)
return s;
else if (r == 2)
{
for (int i = 0; i < len; i += 2)
res += s[i];
for (int j = 1; j < len; j += 2)
res += s[j];
return res;
}
char chess[1000][1000] = {};
for (int i = 0; i < r; i++)
for (int j = 0; j < len; j++)
chess[i][j] = ' ';//初始化

int it = 0, row = 0, col = 0;
while (it < len)
{
chess[row][col] = s[it];
if (row == 0)
row++;
else if (row == r - 1)
{
row--; col++;
}
else if (col % (r - 1) == 0)
row++;
else
{
row--; col++;
}
it++;
}

for (int i = 0; i < r; i++)
for (int j = 0; j <= col; j++)
if (chess[i][j] != ' ')
res += chess[i][j];

return res;
}
};

//29ms
class Solution {
public:
string convert(string s, int r)
{

string result = "";
if (r == 1)
return s;
int step1, step2;
int len = s.size();
for (int i = 0; i<r; ++i)
{
step1 = (r - i - 1) * 2;
step2 = (i) * 2;
int pos = i;
if (pos<len)
result += s[pos];
while (1)
{
pos += step1;
if (pos >= len)
break;
if (step1)
result += s[pos];
pos += step2;
if (pos >= len)
break;
if (step2)
result += s[pos];
}
}
return result;
}

};