[Leetcode] 006. ZigZag Conversion (Easy) (C++/java/python)

Index: [Leetcode] leetcode key index (C++/JAVA/PYTHON/SQL)
Github:https://github.com/illuz/leetcode

006.zigzag_conversion (Easy) links：

Title: https://oj.leetcode.com/problems/zigzag-conversion/
Code (GitHub): Https://github.com/illuz/leetcode

Test Instructions：

Arranges a string in a line that is written in horizontal order.

Analysis：

Direct simulation on the line.

Code: C + +:

Class Solution {Public:string convert (string s, int nRows) {if (nRows = = 1) return s;int step = nRows * 2-2, len = S.len Gth (); 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

[Leetcode] 006. ZigZag Conversion (Easy) (C++/java/python)