6. ZigZag Conversion LeetCode题解
2017-05-10 10:35
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The string
this: (you may want to display this pattern in a fixed font for better legibility)
字符串
And then read line by line:
按行读取,则是
Write the code that will take a string and make this conversion given a number of rows:
return
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题解:
字符串类题目,考虑两种思路:
1. 按照题意方式进行模拟,对于每一行开一个StringBuilder,按照旋转顺序,依次将字符插入到对应行的StringBuilder中,最后将这些StringBuilder拼起来;
2. 直接计算新字符串中下一个字符在原字符中的位置;
代码采取第二种思路,不难发现,zig旋转方向的下一个字符计算公式为k * (n -1 + n - 2 + 1) + line = k * (n * 2 - 2) + line
其中k表示第几列zig,n表示共有几行,line表示当前行数
同样,如果改行存在zag,则公式为(k + 1) * (2 * n - 2) - line
Code【Java】
public class Solution {
private boolean addChar(StringBuilder sb, String s, int pos) {
if (pos < 0 || pos >= s.length()) {
return false;
}
sb.append(s.charAt(pos));
return true;
}
public String convert(String s, int numRows) {
// 处理特殊情况
if (s == null || s.length() <= numRows || numRows < 2) {
return s;
}
StringBuilder sb = new StringBuilder();
for (int i = 0; i < numRows; ++i) {
for (int k = 0; ; ++k) {
// 竖直方向添加字符
int pos1 = k * (2 * numRows - 2) + i;
if (!addChar(sb, s, pos1)) break;
// 倾斜方向添加字符
if (i == 0 || i == numRows - 1) {
continue;
}
int pos2 = (k + 1) * (2 * numRows - 2) - i;
if (!addChar(sb, s, pos2)) break;
}
}
return sb.toString();
}
}
Code【C++】
class Solution {
bool addChar(stringstream& ss, string& s, int pos) {
if (pos < 0 || pos >= s.length()) {
return false;
}
ss << s[pos];
return true;
}
public:
string convert(string s, int numRows) {
// 处理特殊情况
if (s.length() <= numRows || numRows < 2) {
return s;
}
stringstream ss;
for (int i = 0; i < numRows; ++i) {
for (int k = 0; ; ++k) {
// 竖直方向添加字符
int pos1 = k * (2 * numRows - 2) + i;
if (!addChar(ss, s, pos1)) break;
// 倾斜方向添加字符
if (i == 0 || i == numRows - 1) {
continue;
}
int pos2 = (k + 1) * (2 * numRows - 2) - i;
if (!addChar(ss, s, pos2)) break;
}
}
return ss.str();
}
};
"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)
字符串
"PAYPALISHIRING"写成左旋右旋的形式如下所示:
P A H N A P L S I I G Y I R
And then read line by line:
"PAHNAPLSIIGYIR"
按行读取,则是
"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".
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题解:
字符串类题目,考虑两种思路:
1. 按照题意方式进行模拟,对于每一行开一个StringBuilder,按照旋转顺序,依次将字符插入到对应行的StringBuilder中,最后将这些StringBuilder拼起来;
2. 直接计算新字符串中下一个字符在原字符中的位置;
代码采取第二种思路,不难发现,zig旋转方向的下一个字符计算公式为k * (n -1 + n - 2 + 1) + line = k * (n * 2 - 2) + line
其中k表示第几列zig,n表示共有几行,line表示当前行数
同样,如果改行存在zag,则公式为(k + 1) * (2 * n - 2) - line
Code【Java】
public class Solution {
private boolean addChar(StringBuilder sb, String s, int pos) {
if (pos < 0 || pos >= s.length()) {
return false;
}
sb.append(s.charAt(pos));
return true;
}
public String convert(String s, int numRows) {
// 处理特殊情况
if (s == null || s.length() <= numRows || numRows < 2) {
return s;
}
StringBuilder sb = new StringBuilder();
for (int i = 0; i < numRows; ++i) {
for (int k = 0; ; ++k) {
// 竖直方向添加字符
int pos1 = k * (2 * numRows - 2) + i;
if (!addChar(sb, s, pos1)) break;
// 倾斜方向添加字符
if (i == 0 || i == numRows - 1) {
continue;
}
int pos2 = (k + 1) * (2 * numRows - 2) - i;
if (!addChar(sb, s, pos2)) break;
}
}
return sb.toString();
}
}
Code【C++】
class Solution {
bool addChar(stringstream& ss, string& s, int pos) {
if (pos < 0 || pos >= s.length()) {
return false;
}
ss << s[pos];
return true;
}
public:
string convert(string s, int numRows) {
// 处理特殊情况
if (s.length() <= numRows || numRows < 2) {
return s;
}
stringstream ss;
for (int i = 0; i < numRows; ++i) {
for (int k = 0; ; ++k) {
// 竖直方向添加字符
int pos1 = k * (2 * numRows - 2) + i;
if (!addChar(ss, s, pos1)) break;
// 倾斜方向添加字符
if (i == 0 || i == numRows - 1) {
continue;
}
int pos2 = (k + 1) * (2 * numRows - 2) - i;
if (!addChar(ss, s, pos2)) break;
}
}
return ss.str();
}
};
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