LeetCode题解 第八周

1.ZigZag Conversion

https://leetcode.com/problems/zigzag-conversion/description/

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".

Difficulty：Medium

Explanation：

1.这道题目的解法核心在于找到字母拍成z字形的规律。当一行字符按照zigzag规则排列时，其实是满足一定周期性的，每个周期就可以看作是一个竖行和一个斜着的行，然后可以据此找到元素坐标和在z字中的位置的函数。但是要注意当设置的z字行数小于3的时候这个规律就不适用了，要另外考虑。

code：

class Solution {public:    string convert(string s, int numRows) {vector<string> ZigZag;for (int i = 0; i < numRows; i++){ZigZag.push_back("");}if (numRows <= 2){for (int i = 0; i < s.size(); i++){ZigZag[i%numRows] += s[i];}}else{int nums = numRows - 2;for (int i = 0; i < s.size(); i++){int GroupNum = (i + 1) % (nums + numRows);if (GroupNum <= numRows&&GroupNum>0){ZigZag[GroupNum - 1] += s[i];}else if (GroupNum == 0){ZigZag[1] += s[i];}else{ZigZag[numRows - (GroupNum - numRows) - 1] += s[i];}}}string temp="";for (int i = 0; i < numRows; i++){temp += ZigZag[i];}return temp;}};