Leetcode-6. ZigZag Conversion

来源：互联网发布：gg是什么意思网络用语编辑：程序博客网时间：2024/04/24 20:38

前言：为了后续的实习面试，开始疯狂刷题，非常欢迎志同道合的朋友一起交流。因为时间比较紧张，目前的规划是先过一遍，写出能想到的最优算法，第二遍再考虑最优或者较优的方法。如有错误欢迎指正。博主首发CSDN，mcf171专栏。

博客链接：mcf171的博客

——————————————————————————————

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   NA P L S I I GY   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".

第一个思路是直接按照Z格式先写了，然后用O（n^2）的办法得到结果。Your runtime beats 5.40% of java submissions

public class Solution {    public String convert(String s, int numRows) {        if(numRows == 1)        return s;        int numColumns = (s.length()/(numRows - 1 )) ;        numColumns = numColumns == 0 ? 1 : numColumns;        char[][] result = new char[numRows][numColumns];        boolean direction = true;        int row = 0, column = 0;        for(int i = 0 ; i < s.length(); i ++){            result[row][column] = s.charAt(i);            if(direction == true){                row ++;                if(row == (numRows - 1)) direction = false;            }else{                row --;                column ++;                if(row == 0) direction = true;            }        }        StringBuffer bf = new StringBuffer("");        for(int i = 0 ;i < numRows ; i ++)            for(int j = 0 ; j < numColumns; j ++){                if('\0' != result[i][j]) bf.append(result[i][j]);            }        return bf.toString();    }}

第二个思路是采用跳着访问矩阵的方式输出结果。Your runtime beats 23.41% of java submissions.

public class Solution {    public String convert(String s, int numRows) {        if(numRows == 1)            return s;        int numColumns = (s.length()/(2*numRows - 2 ) + 1) * (1 + numRows -2);        char[][] result = new char[numRows][numColumns];        boolean direction = true;        int row = 0, column = 0;        for(int i = 0 ; i < s.length(); i ++){            result[row][column] = s.charAt(i);            if(direction == true){                row ++;                if(row == (numRows - 1)) direction = false;            }else{                row --;                column ++;                if(row == 0) direction = true;            }        }        StringBuffer bf = new StringBuffer("");        for(int i = 0 ;i < numRows ; i ++) {            boolean jump = true;            for (int j = 0; j < numColumns;) {                if ('\0' != result[i][j]) bf.append(result[i][j]);                if( i == 0 || i == numRows -1)                    j = j + numRows - 1;                else{                    if(jump) {                        j = j + numRows - 1 - i;                    }else{                        j = j + i;                    }                    jump = !jump;                }            }        }        return bf.toString();    }}

第三个思路，发现其实每个字符串跳跃是有规律的，都不用矩阵了，直接跳跃就好了，复杂度O(n)Your runtime beats 32.95% of java submissions.

public class Solution {    public String convert(String s, int numRows) {        if(numRows == 1)            return s;        StringBuffer bf = new StringBuffer("");        for(int i = 0 ;i < numRows ; i ++) {            boolean jump = true;            int index = i;    for(;index < s.length();){    bf.append(s.charAt(index));            if( i == 0 || i == numRows -1)                index += numRows * 2 -2 ;            else{                if(jump) {                   index += numRows * 2 - 2 - ( i * 2);                }else{                   index += i* 2;                }                jump = !jump;                }    }        }        return bf.toString();    }}

0 0