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LeetCode

Approach: Sort by Row, Time Complexity: O(n), where n == len(s)

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class Solution {
    public String convert(String s, int numRows) {
        if(numRows==1) return s;
        List<StringBuilder> rows=new ArrayList<>();
        for(int i=0;i<Math.min(numRows,s.length());i++) rows.add(new StringBuilder());
        
        int curRow=0;
        boolean goingDown=false;
        for(char ch:s.toCharArray()){
            rows.get(curRow).append(ch);
            if(curRow==0||curRow==numRows-1) goingDown=!goingDown;
            curRow+=goingDown?1:-1;
        }
        
        StringBuilder sb=new StringBuilder();
        for(StringBuilder row:rows) sb.append(row);
        return sb.toString();
    }
}

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