LeetCode 题解(240) : Strobogrammatic Number III

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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.

题解：

在Strobogrammatic Number II的基础上，统计从len(low)到len(high)为n的所有在范围内的个数。Python会超时就不给出了。

首先给出无优化版本，296ms。

class Solution { public:     int strobogrammaticInRange(string low, string high) {         this->low = low;         this->high = high;         unordered_map<char, string> d;         d.insert(pair<char, string>('0', "0"));         d.insert(pair<char, string>('1', "1"));         d.insert(pair<char, string>('6', "9"));         d.insert(pair<char, string>('8', "8"));         d.insert(pair<char, string>('9', "6"));                  string s = "01869";          for(int i = low.length(); i <= high.length(); i++) {             int mid = i / 2;             string result = "";             bool oneMore = (bool)(i % 2);             recursion(d, result, mid, 0, s, oneMore);         }         return this->count;     }          void recursion(unordered_map<char, string> &d, string result, int recur, int iter, string s, bool &oneMore) {         if(iter > recur) {             for(int i = 0; i < 3; i++) {                 if(stoll(result.substr(0, recur) + s[i] + result.substr(recur)) <= stoll(this->high) && stoll(result.substr(0, recur) + s[i] + result.substr(recur)) >= stoll(this->low)) {                     this->count++;                 }             }             return;         }         if(iter == recur) {             if(oneMore) {                 recursion(d, result, recur, iter + 1, s, oneMore);             } else {                 if(stoll(result) <= stoll(this->high) && stoll(result) >= stoll(this->low)) {                     this->count++;                 }             }             return;         }         if(iter == 0) {             for(int i = 1; i < 5; i++) {                 recursion(d, s[i] + d[s[i]], recur, iter + 1, s, oneMore);             }         } else {             for(int i = 0; i < 5; i++) {                 recursion(d, result.substr(0, iter) + s[i] + d[s[i]] + result.substr(iter), recur, iter + 1, s, oneMore);             }         }     }      private:     string low, high;     int count = 0; };

后来想想，可以把之前的计算结果缓存起来，这样不用每次重新计算。然并卵。还是需要196ms，只提升了100ms。

class Solution {public:    int strobogrammaticInRange(string low, string high) {        this->low = stoll(low);        this->high = stoll(high);        unordered_map<char, string> d;        d.insert(pair<char, string>('0', "0"));        d.insert(pair<char, string>('1', "1"));        d.insert(pair<char, string>('6', "9"));        d.insert(pair<char, string>('8', "8"));        d.insert(pair<char, string>('9', "6"));                unordered_map<int, vector<string>> save;                string s = "01869";        for(int i = low.length(); i <= high.length(); i++) {            vector<string> results;            int mid = i / 2;            string result = "";            bool oneMore = (bool)(i % 2);            if(oneMore && save.find(i - 1) != save.end()) {                for(auto j : save[i - 1]) {                    recursion(d, j, mid, mid, s, oneMore, results);                }            } else if(!oneMore && save.find(i - 2) != save.end()) {                for(auto j : save[i - 2]) {                    recursion(d, j, mid, mid - 1, s, oneMore, results);                }            } else {                recursion(d, "", mid, 0, s, oneMore, results);            }            save.insert(pair<int, vector<string>>(i, results));        }        return this->count;    }        void recursion(unordered_map<char, string> &d, string result, int recur, int iter, string s, bool &oneMore, vector<string> &results) {        if(iter > recur) {            for(int i = 0; i < 3; i++) {                string t = result.substr(0, recur) + s[i] + result.substr(recur);                long long r = stoll(t);                if(r <= this->high && r >= this->low) {                    this->count++;                    results.push_back(t);                }            }            return;        }        if(iter == recur) {            if(oneMore) {                recursion(d, result, recur, iter + 1, s, oneMore, results);            } else {                long long r = stoll(result);                if(r <= this->high && r >= this->low) {                    this->count++;                    results.push_back(result);                }            }            return;        }        if(iter == 0) {            for(int i = 1; i < 5; i++) {                recursion(d, s[i] + d[s[i]], recur, iter + 1, s, oneMore, results);            }        } else {            for(int i = 0; i < 5; i++) {                recursion(d, result.substr(0, iter) + s[i] + d[s[i]] + result.substr(iter), recur, iter + 1, s, oneMore, results);            }        }    }    private:    long long low, high;    int count = 0;};

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