248. Strobogrammatic Number III

题目：

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

Write a function to count the total strobogrammatic numbers that exist in the range of low <= num <= high.

For example,
Given low = "50", high = "100", return 3. Because 69, 88, and 96 are three strobogrammatic numbers.

Note:
Because the range might be a large number, the low and high numbers are represented as string.

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

题解：

由于有了上一题目的臭长解法，这里继续沿用...优化留给二刷吧，看官图个乐子好了。 思路是使用Strobogrammatic Number II的代码，我们对于从 low.length()到high.length()的每一个n，求解Strobogrammatic Number，然后把这个数字与low和high进行比较，假如这个数字在[low，high]的闭区间里，则可以算作一个结果。下面的方法大量使用了Strobogrammatic Number II的代码，其实有很多地方可以优化，比如不需要传递一个List<>，只用传递一个数组，用首元素进行计数就可以了，比如int[] arr = new int[1]，然后用a[0]来记录数字的增减。 还有methods可以由四个变为三个，合并第一和第二个method，这样也可以不用每次都create一个map，等等。复杂度也要好好计算一下。

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

public class Solution {
    public int strobogrammaticInRange(String low, String high) {
        if(low == null || high == null)
            return 0;
        int lo = low.length(), hi = high.length();
        int res = 0;
 
        for(int i = lo; i <= hi; i++)
            res += findStrobogrammatic(i, low, high).size();
 
        return res;
    }
 
    private List<String> findStrobogrammatic(int n, String low, String high) {
        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, low, high);
 
        return res;
    }
 
    private void findStrobogrammatic(List<String> res, StringBuilder sb, Map<Character, Character> map, int n, int position, String low, String high) {
        if(sb.length() > n)
            return;
        if(sb.length() == n) {
            String s = sb.toString();
            if(firstStringEqualToOrSmaller(low, s) && firstStringEqualToOrSmaller(s, high))
                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, low, high);
                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, low, high);
                sb.deleteCharAt(0);
                sb.deleteCharAt(sb.length() - 1);
            }
        }
    }
 
    private boolean firstStringEqualToOrSmaller(String s, String t) {
        if(s.length() < t.length())
            return true;
        else if(s.length() > t.length())
            return false;
        else {
            for(int i = 0; i < s.length(); i++)
                if(s.charAt(i) > t.charAt(i))
                    return false;
                else if(s.charAt(i) < t.charAt(i))
                    return true;
            return true;
        }
    }
}

Reference:

https://leetcode.com/discuss/55468/clear-java-ac-solution-using-strobogrammatic-number-method

https://leetcode.com/discuss/54562/clean-and-easy-java-recursive-solution

https://leetcode.com/discuss/50628/ac-java-solution-with-explanation

https://leetcode.com/discuss/50624/clean-and-easy-understanding-java-solution

https://leetcode.com/discuss/50604/solution-based-on-strobogrammatic-number-ii