Letter Combinations of a Phone Number (LeetCode)

Question:

https://oj.leetcode.com/problems/letter-combinations-of-a-phone-number/

 

解答：

Combination的问题，一般就是DFS和BFS枚举所有的情况。

BFS好像只有iterative的解法比较普遍，recursive的解法不多见。

DFS可以iterative，也可以recursive.

Iterative解法DFS用stack, BFS用queue。 注意stack取值是stack.top(), queue是queue.front().

class Solution {
public:

    vector<string> letterCombinations(string digits) {
         std::map<char, string> keyboard;
        keyboard['0'] = " ";
        keyboard['2'] = "abc";
        keyboard['3'] = "def";
        keyboard['4'] = "ghi";
        keyboard['5'] = "jkl";
        keyboard['6'] = "mno";
        keyboard['7'] = "pqrs";
        keyboard['8'] = "tuv";
        keyboard['9'] = "wxyz";
        
        vector<string> ret;

        //std::stack<std::pair<int, string>> container;
        std::queue<std::pair<int, string>> container;
        std::pair<int, string> sp = std::make_pair(0, "");
        container.push(sp);
        
        while (!container.empty())
        {
            //sp = container.top();
            sp = container.front();
            container.pop();
            if (sp.first == digits.size())
            {
                ret.push_back(sp.second);
            }
            else
            {
                for (int i = 0; i < keyboard[digits[sp.first]].size(); i++)
                {
                    std::pair<int, string> newSp(sp);
                    newSp.first++;
                    newSp.second.push_back(keyboard[digits[sp.first]][i]);
                    container.push(newSp);
                }
            }
        }
        
        return ret;
    }
};