[Leetcode] 248. Strobogrammatic Number III 解题报告

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.

思路：

在《[Leetcode] 247. Strobogrammatic Number II 解题报告》的基础上稍作改进就可以了。也就是枚举从low的长度到high的长度之间的符合条件的数，一旦发现它在low和high的范围之内，就更新计数器。然而，总觉得这种解法不是最优的，因为从直观上来说我们可能不需要枚举所有符合条件的数。等后面有空再想想具体做法吧。

代码：

class Solution {public:    int strobogrammaticInRange(string low, string high) {        int n1 = low.length(), n2 = high.length();        int ret = 0;        for (int i = n1; i <= n2; ++i) {            dfs(low, high, i, "", ret);        }        return ret;    }private:    void dfs(string &low, string &high, int n, string str, int &result) {          if(n == 0) {              if (stol(str) >= stol(low) && stol(str) <= stol(high)) {                ++result;            }            return;        }          if(n % 2 != 0) {            for(auto val : same) {                  dfs(low, high, n-1, val, result);              }          }          else {              for(int i = (n == 2) ? 1 : 0; i < two.size(); ++i) {                  dfs(low, high, n-2, two[i].first + str + two[i].second, result);              }          }      }      vector<string> same{"0", "1", "8"};        vector<pair<char,char>> two{{'0','0'},{'1','1'},{'6','9'},{'8','8'},{'9','6'}};};