248. Strobogrammatic Number III:

其實(shí)抽絲剝繭，這題就是一道recursion的題目，先去不停的構(gòu)建在長(zhǎng)度范圍內(nèi)的值，然后選取合適的值count一下

class Solution(object):
    def strobogrammaticInRange(self, low, high):
        """
        :type low: str
        :type high: str
        :rtype: int
        """
        count = 0
        rst = []
        for i in range(len(low), len(high)+1):
            rst += self.helper(i)
        for num in rst:
            if (len(num) == len(low) and num < low) or (len(num) == len(high) and num > high) or (num[0] == "0" and num != "0"):
                continue
            count += 1
        return count
    
    def helper(self, cur):
        if cur <= 0:
            return [""]
        if cur == 1:
            return ["1","8","0"]
        rst = []
        center = self.helper(cur-2);
        
        for i in range(len(center)):
            tmp = center[i]
            rst.append("0" + tmp + "0")
            rst.append("1" + tmp + "1")
            rst.append("6" + tmp + "9")
            rst.append("8" + tmp + "8")
            rst.append("9" + tmp + "6")
        return rst