leetcode [#6]

目录

• 1. 题目
• 2. 解决方案
• 3. 注意事项

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”.

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解决方案

public class Solution {
    public String convert(String s, int numRows) {
        int len = s.length();
        if(len == 0) return "";
        if(numRows == 1) return s;
        char[] sArr = s.toCharArray();
        StringBuilder[] b = new StringBuilder[numRows];
        for(int k = 0; k < numRows; k++){
            b[k] = new StringBuilder();
        }
        int piece = 2 * (numRows - 1);
        for(int i = 0; i < len; i++){
            if(i < numRows){
                b[i].append(sArr[i]);
            } else {
                if (i % piece >= 0 && i % piece < numRows) {
                    b[i % piece].append(sArr[i]);
                } else {
                    b[piece - i % piece].append(sArr[i]);
                }
            }
        }
        StringBuilder resBuilder = new StringBuilder();
        for(int m = 0; m < numRows; m++){
            resBuilder.append(b[m]);
        }
        return  resBuilder.toString();
    }
}

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注意事项

1. 按照给定的numRows，将字符串排列成锯齿状，并且按照行返回。
2. 先处理特殊情况。
3. 定义numRows个StringBuilder，重点在于找到每个字符应该加入的StringBuilder。对于下标满足i % piece >= 0 && i % piece < numRows范围的字符串，即锯齿中“垂直方向”的一段，依次填入b[i % piece]即可；对于下标满足i % piece >= numRows范围的字符串，即锯齿中“倾斜方向”的一段，依次填入b[piece - i % piece]即可。