Leetcode 6. ZigZag Conversion

这道题比较简单，观察清楚规律后很快可以写出来。

但是用go写的版本，总是答案错误，本地无问题。后用Java重写。

func convert(s string, numRows int) string {
    if numRows == 1 {
        return s
    }
    numChars := make(map[int][]string)
    size := 2 * numRows - 2
    for i := 0; i<len(s); i++ {
        x := s[i : i+1]
        if i % size < numRows {
            row := i % size
            numChars[row] = append(numChars[row], x)
        } else {
            row := size - i % size
            numChars[row] = append(numChars[row], x)
        }
    }

    var result string

    for _, ss := range numChars {
        for _, c := range ss {
            result += c
        }
    }

    return result
}

class Solution {
    public String convert(String s, int numRows) {
        if (numRows == 1 || s == null) {
            return s;
        }
        Map<Integer, List<String>> numChars = new HashMap<Integer, List<String>>();
        int size = 2 * numRows - 2;
        for (int i = 0; i < s.length(); i++) {
            String x = s.substring(i, i + 1);
            int row;
            if (i % size < numRows) {
                row = i % size;
            } else {
                row = size - i % size;
            }
            if (numChars.get(row) == null) {
                numChars.put(row, new ArrayList<String>());
            }
            numChars.get(row).add(x);
        }

        StringBuilder result = new StringBuilder();
        for (List<String> ss : numChars.values()) {
            for (String c : ss) {
                result.append(c);
            }
        }

        return result.toString();
    }
}

Leetcode 6. ZigZag Conversion

原文：https://www.cnblogs.com/yx1989/p/10986619.html