[LeetCode] 006. ZigZag Conversion (Easy) (C++/Java/Python)，leetcodezigzag

[LeetCode] 006. ZigZag Conversion (Easy) (C++/Java/Python)，leetcodezigzag

索引：[LeetCode] Leetcode 題解索引 (C++/Java/Python/Sql)
Github: https://github.com/illuz/leetcode

006.ZigZag_Conversion (Easy) 連結：

題目：https://oj.leetcode.com/problems/zigzag-conversion/
代碼(github)：https://github.com/illuz/leetcode

題意：

把一個字串按橫寫的折線排列。

分析：

直接類比就行了。

代碼：C++:

class Solution {public:string convert(string s, int nRows) {if (nRows == 1) return s;int step = nRows * 2 - 2, len = s.length();string ret = "";// first rowfor (int i = 0; i < len; i += step)ret += s[i];for (int i = 1; i < nRows - 1; i++) {for (int j = i; j < len; j += step) {ret += s[j];if (j + (step - i * 2) < len)ret += s[j + (step - i * 2)];}}// last rowfor (int i = nRows - 1; i < len; i += step)ret += s[i];return ret;}};

Java:

public class Solution {    public String convert(String s, int nRows) {        if (nRows == 1) return s;        int step = nRows * 2 - 2, len = s.length();        String ret = "";        // first row        for (int i = 0; i < len; i += step)            ret += s.charAt(i);        for (int i = 1; i < nRows - 1; i++) {            for (int j = i; j < len; j += step) {                ret += s.charAt(j);                if (j + (step - i * 2) < len)                    ret += s.charAt(j + (step - i * 2));            }        }        // last row        for (int i = nRows - 1; i < len; i += step)            ret += s.charAt(i);        return ret;    }}

Python:

class Solution:    # @return a string    def convert(self, s, nRows):        if nRows == 1:            return s        step = nRows * 2 - 2        # first row        ret = s[::step]        for i in range(1, nRows - 1):            for j in range(i, len(s), step):                ret += s[j]                if j + (step - i * 2) < len(s):                    ret += s[j + (step - i * 2)]        # last row        ret += s[nRows - 1::step]        return ret