leetcode题解-6. ZigZag Conversion

题意：将字符串按照锯齿形排列，然后把从上到下把所有行顺序输出，组成新的字符串。

例子：
Input: s = “PAYPALISHIRING”, numRows = 3
Output: “PAHNAPLSIIGYIR”

Example 2:

Input: s = “PAYPALISHIRING”, numRows = 4
Output: “PINALSIGYAHRPI”
Explanation:

分析：对于每一行先把往下走的那一列的字符加进去，然后有往上走的字符再加进去即可。时间复杂度是O(n),空间复杂度是O(n),代码如下：

class Solution {
    public String convert(String s, int numRows) {
        if(numRows == 1) return s;
        ArrayList<StringBuilder> sbArr = new ArrayList<>();
        for(int i = 0; i < Math.min(numRows, s.length()); i++){
            StringBuilder sb = new StringBuilder();
            sbArr.add(sb);
        }
        // 用来统计当前是第几行
        int rowCount = 0;
        // 用来统计是向下查还是向上查
        int downCount = 1;

        for(int i = 0; i < s.length(); i++){
            if(downCount == 1){
                sbArr.get(rowCount).append(s.charAt(i));
                if(rowCount < numRows - 1){
                    rowCount++;
                }else{
                    downCount = 1 - downCount;
                    rowCount--;
                }
            }else{
                sbArr.get(rowCount).append(s.charAt(i));
                if(rowCount > 0){
                    rowCount--;
                }else{
                    downCount = 1 - downCount;
                    rowCount++;
                }
            }
        }
        StringBuilder result = new StringBuilder();
        for(StringBuilder cStr : sbArr){
            result.append(cStr);
        }
        return result.toString();

    }
}