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Commit 62dfbd3

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‎6. ZigZag Conversion.java

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/*
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题目:
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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)
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P A H N
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A P L S I I G
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Y I R
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And then read line by line: "PAHNAPLSIIGYIR"
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Write the code that will take a string and make this conversion given a number of rows:
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string convert(string text, int nRows);
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convert("PAYPALISHIRING", 3) should return "PAHNAPLSIIGYIR".
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*/
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/*
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解析:
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这道题我是没有思路的,完全不知道怎么做,想过用数组存储,但是还是不太清楚。
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后来看过别人的思路,发现这种题,就相当于找规律,其实和他的那个之字形没多大关系
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先看一下:假设行数为4
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0 6 12
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1 5 7 11 13
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2 4 8 10 14
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3 9 15
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这就是所谓的之字形排布,
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假设行数为row(从0开始)那么每行的间隔:
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第0行和最后一行:总间隔 = 2row(2*3=6)
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其余行:第一个元素与第二个元素的间隔:总间隔-2row 第二个元素与第三个元素:2row 之后依次类推。。。
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找出这个规律之后,就不难了。。
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需要注意的是,需要用一个flag来记录,其余行中不同间隔的转换,每轮循环别忘了恢复flag的值。
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不懂得话,具体看代码:
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*/
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class Solution {
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public String convert(String s, int numRows) {
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int flag = 1;
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int row = 0; //行数从0开始计数。
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int confirm = 2*(numRows-1);
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int part = confirm;//设置间隔,初始是2(行数-1)
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//新建一个结果字符串
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String result = "";
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//针对一些特殊的情况
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//行数为1
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if(numRows==1) {
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return s;
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}
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while(row<numRows) {
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//第0行和最后一行,特殊照顾!
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if(row==0||row==numRows-1) {
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for(int i = row;i<s.length();i+=confirm) {
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result += s.charAt(i);
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}
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row++;
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}
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//其余行
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else {
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for(int i = row ;i<s.length();i+=part) {
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result +=s.charAt(i);
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if(flag==1){
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part=confirm-2*row;
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flag=0;
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}
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else {
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part=2*row;
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flag=1;
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}
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}
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row++;//行数++
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flag=1;//每次要恢复flag的值
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}
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}
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return result;
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}
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}

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