6-Z字形变换
将一个给定字符串根据给定的行数,以从上往下、从左到右进行 Z 字形排列。
比如输入字符串为 "LEETCODEISHIRING" 行数为 3 时,排列如下:
L C I R
E T O E S I I G
E D H N
之后,你的输出需要从左往右逐行读取,产生出一个新的字符串,比如:"LCIRETOESIIGEDHN"。
请你实现这个将字符串进行指定行数变换的函数:
string convert(string s, int numRows);
示例 1:
输入: s = "LEETCODEISHIRING", numRows = 3
输出: "LCIRETOESIIGEDHN"
示例 2:
输入: s = "LEETCODEISHIRING", numRows = 4
输出: "LDREOEIIECIHNTSG"
解释:
L D R
E O E I I
E C I H N
T S G
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/zigzag-conversion
著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
public String convert(String s, int numRows) {
if(s == null || s.equals("") || numRows == 1) {
return s;
}
int numCols = (s.length() / (numRows * 2 - 2) + 1) * (numRows - 1);
char[][] chs = new char[numRows][numCols];
int i = 0;
int j = 0;
char[] ss = s.toCharArray();
chs[i][j] = ss[0];
for (int k = 1; k < ss.length; ) {
// 往下填充
while (i < numRows - 1 && k < ss.length) {
chs[++i][j] = ss[k++];
}
// 左下到右上
while (i > 0 && k < ss.length) {
chs[--i][++j] = ss[k++];
}
}
StringBuilder sb = new StringBuilder();
for (char[] c : chs) {
for (char ch : c) {
if (ch != 0) {
sb.append(ch);
}
}
}
return sb.toString();
}
找规律
public String convert(String s, int numRows) {
if(s == null || s.equals("") || numRows == 1) {
return s;
}
int step = numRows * 2 - 2;
char[] ss = s.toCharArray();
StringBuilder sb = new StringBuilder();
// i = 0
for(int i = 0; i < s.length(); i += step) {
sb.append(ss[i]);
}
// i = 1 | i = step-1
for(int k = 1; k < numRows - 1; k++) {
int j = step - k;
for (int i = k; i < s.length(); i += step, j += step) {
sb.append(ss[i]);
if(j < s.length()) {
sb.append(ss[j]);
}
}
}
// i = numRows-1
for(int i = numRows - 1; i < s.length(); i += step) {
sb.append(ss[i]);
}
return sb.toString();
}
官方题解
class Solution {
public String convert(String s, int numRows) {
if (numRows == 1) return s;
List<StringBuilder> rows = new ArrayList<>();
for (int i = 0; i < Math.min(numRows, s.length()); i++)
rows.add(new StringBuilder());
int curRow = 0;
boolean goingDown = false;
for (char c : s.toCharArray()) {
rows.get(curRow).append(c);
if (curRow == 0 || curRow == numRows - 1) goingDown = !goingDown;
curRow += goingDown ? 1 : -1;
}
StringBuilder ret = new StringBuilder();
for (StringBuilder row : rows) ret.append(row);
return ret.toString();
}
}
作者:LeetCode
链接:https://leetcode-cn.com/problems/zigzag-conversion/solution/z-zi-xing-bian-huan-by-leetcode/
来源:力扣(LeetCode)
著作权归作者所有。商业转载请联系作者获得授权,非商业转载请注明出处。