6-e-ZigZag Conversion

好久没刷leet了，最近想着把以前遗留的题重新刷一下。

这道折线排列字符的题目记得是刚开始刷leet的时候，对leet的难度和规律还没有映像，当时觉得很难而且理解还有误区（英语理解不准确），导致写了好大一段代码最后还是deny。

现在重新看了一下，发现其实是个数学题，大致规律就是相邻两竖线中同列两字符的差值一定等于(numRows - 1)*2，如果其两字符中间有间隔字符的话，这些间隔字符的间距和一定等于(numRows - 1)*2，有点类似上半边三角形+下半边三角形=整三角形。找出了数学规律再编程就不难了。

贴出AC代码一目了然：

char *fixed_convert(char* s, int numRows) {if (numRows == 1)return s;char *result = NULL;int length = (int)strlen(s);result = (char *)calloc(length + 1, sizeof(char));int strIndex = 0, r_column = 0, r_index = 0;bool firstIn = true;for (int i = 0; i < numRows; i++) {r_index = 0, firstIn = true;while (strIndex < length) {if (firstIn) {r_column = i;firstIn = false;}else if ( ((r_index % 2 == 1)||(i == 0)) && (i < numRows - 1) )r_column += (numRows - i - 1) * 2;else if (r_index % 2 == 0 || i == numRows - 1)r_column += i * 2;if (r_column >= length)break;result[strIndex] = s[r_column];strIndex++;r_index++;}}result[length] = '\0';return result;
}

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