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"
.
思路:
该题属于模拟范畴,不难,但是需要注意边界的条件细节。
找出string下标和所在行数的关系。其中str数组需要能够随机访问到,这样在查找的时候可以直接根据下标关系在O(1)的时间复杂度中找到。
AC代码:
public String convert(String s, int nRows) { if(nRows==1) return s; char[] strs = s.toCharArray(); LinkedList<Integer> head = new LinkedList<Integer>(); head.add(0); for(int i=1;;i++){ int h = i*(2*nRows-2); if(h<strs.length) head.add(h); else{ head.add(h); break; } } StringBuilder sb = new StringBuilder(""); for(int i=0;i<nRows;i++){ for(Integer idx:head){ if(idx-i>=0 && idx-i<idx && idx-i<strs.length && i<nRows-1){ //此处条件略复杂,针对重复下标和越界的判断进行处理 sb.append(strs[idx-i]); } if(idx+i>=idx && idx+i<strs.length) //<span style="font-family: 'Helvetica Neue', Helvetica, Arial, sans-serif;">此处条件略复杂,针对重复下标和越界的判断进行处理</span> sb.append(strs[idx+i]); } } return sb.toString(); }