247. Strobogrammatic Number II

题目：

A strobogrammatic number is a number that looks the same when rotated 180 degrees (looked at upside down).

Find all strobogrammatic numbers that are of length = n.

For example,
Given n = 2, return ["11","69","88","96"].

Hint:

Try to use recursion and notice that it should recurse with n - 2 instead of n - 1.

链接： http://leetcode.com/problems/strobogrammatic-number-ii/

题解：

求所有长度为n的strobogrammatic number。这题一开始的思路是用DFS + Backtracking。还要处理一些特殊的边界条件，比如n的长度为奇数和偶数，以及最外层不能为两个'0'等等，代码写得很拖沓。 Discuss区有不少好的代码，二刷时一定要思考清楚再进行优化。这里我主要是从中心向两边添加，而discuss区大家大部分都是从两边向中心递归，所以我的代码还需要回溯，不够简练。

Time Complexity - O(2ⁿ)， Space Complexity - O(n)。

public class Solution {
    public List<String> findStrobogrammatic(int n) {
        if(n < 1)
            return new ArrayList<String>();
        List<String> res = new ArrayList<>();
        Map<Character, Character> map = new HashMap<>();
        map.put('0', '0');
        map.put('1', '1');
        map.put('6', '9');
        map.put('8', '8');
        map.put('9', '6');
        
        StringBuilder sb = new StringBuilder();
        int position = (n % 2 == 0) ? 0 : 1;
        findStrobogrammatic(res, sb, map, n, position);
        
        return res;
    }
    
    private void findStrobogrammatic(List<String> res, StringBuilder sb, Map<Character, Character> map, int n, int position) {
        if(sb.length() > n)
            return;
        if(sb.length() == n) {
            res.add(sb.toString());
            return;
        }
        
        if(position == 1) {
            for(char c : map.keySet()) {
                if(c == '6' || c == '9')
                    continue;
                sb.append(c);
                findStrobogrammatic(res, sb, map, n, position + 1);
                sb.setLength(0);
            }
        } else {
            for(char c : map.keySet()) {
                if(n - sb.length() == 2 && c == '0')
                    continue;
                sb.insert(0, c);
                sb.append(map.get(c));
                findStrobogrammatic(res, sb, map, n, position + 2);
                sb.deleteCharAt(0);
                sb.deleteCharAt(sb.length() - 1);
            }    
        }
    }
}

二刷:

两种思路:

和一刷一样，用dfs + backtracking
求一半String的permutation，剪去一些invalid case，再补上另外一半。也就是使用跟267. Palindrome Permutation II类似的方法。

Java:

Reference:

https://leetcode.com/discuss/50412/ac-clean-java-solution

https://leetcode.com/discuss/50377/my-concise-java-solution-using-dfs

https://leetcode.com/discuss/52277/accepted-java-solution-using-recursion

https://leetcode.com/discuss/53144/my-concise-iterative-java-code

https://leetcode.com/discuss/68215/simple-java-solution-without-recursion

来源：https://www.cnblogs.com/yrbbest/p/5008994.html

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