247. Strobogrammatic Number II输出所有对称数字

[抄题]:

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.

Example:

Input:  n = 2
Output: ["11","69","88","96"]

 [暴力解法]:

时间分析:

空间分析:

 [优化后]:

时间分析:

空间分析:

[奇葩输出条件]:

[奇葩corner case]:

当前长度是0或1时需要退出或者返回,所以要新建一个动态数组。

new ArrayList<String>(Arrays.asList(""));

 

[思维问题]:

不知道怎么控制dfs的长度:

 

helper(curCount - 2, targetCount)取上一截,然后i取list的size

 

[英文数据结构或算法,为什么不用别的数据结构或算法]:

[一句话思路]:

[输入量]:空: 正常情况:特大:特小:程序里处理到的特殊情况:异常情况(不合法不合理的输入):

[画图]:

[一刷]:

curcount != targetcount时,两端必加0,从中间开始加。

[二刷]:

[三刷]:

[四刷]:

[五刷]:

  [五分钟肉眼debug的结果]:

[总结]:

 

helper(curCount - 2, targetCount)取上一截,然后逐渐往外扩展

 

[复杂度]:Time complexity: O(n) Space complexity: O(n)

[算法思想:迭代/递归/分治/贪心]:

[关键模板化代码]:

增长回文串也就只有这几种情况:

 

        res.add("0" + s + "0");
        res.add("1" + s + "1");
        res.add("6" + s + "9");
        res.add("8" + s + "8");
        res.add("9" + s + "6");    

 

[其他解法]:

[Follow Up]:

[LC给出的题目变变变]:

 [代码风格] :

 [是否头一次写此类driver funcion的代码] :

 [潜台词] :

 

class Solution {
    public List<String> findStrobogrammatic(int n) {
        List<String> result = new ArrayList<String>();
        //corner case
        if (n < 0) return result;
        //return
        return getStrobogrammaticNums(n, n);
    }
    
    public List<String> getStrobogrammaticNums(int curCount, int targetCount) {
        //corner case: n = 0 or 1
        if (curCount == 0) return new ArrayList<String>(Arrays.asList(""));
        if (curCount == 1) return new ArrayList<String>(Arrays.asList("0", "1", "8"));
        
        List<String> result = new ArrayList<String>();
        List<String> list = getStrobogrammaticNums(curCount - 2, targetCount);
        
        //get the new nums into result
        for (int i = 0; i < list.size(); i++) {
            String cur = list.get(i);
            //add 0 from inside if length do not equal
            if (curCount != targetCount) result.add("0" + cur + "0");
            //add other nums
            result.add("1" + cur + "1");
            result.add("6" + cur + "9");
            result.add("9" + cur + "6");
            result.add("8" + cur + "8");
        }
        
        //return
        return result;
    }
}
View Code

 

posted @ 2018-08-11 17:36  苗妙苗  阅读(278)  评论(0编辑  收藏  举报