1. Letter Combinations of a Phone Number(17)

题目：

给定一个仅包含数字 2-9 的字符串，返回所有它能表示的字母组合。

给出数字到字母的映射如下（与电话按键相同）。注意 1 不对应任何字母。  

class Solution 
{
    public:
    vector<string> letterCombinations(string digits) 
    {
        vector<string> result;
        if(digits.empty()) return vector<string>();
        static const vector<string> v = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
        result.push_back("");   
        for(int i = 0 ; i < digits.size(); ++i) 
        {
            //得出每一个数字
            int num = digits[i]-'0';
                if(num < 0 || num > 9) break;
            const string& candidate = v[num];
            if(candidate.empty()) 
                continue;
            vector<string> tmp;
            for(int j = 0 ; j < candidate.size() ; ++j) 
            {
                for(int k = 0 ; k < result.size() ; ++k) 
                {
                tmp.push_back(result[k] + candidate[j]);
                }
            }
        result.swap(tmp);
        }
    return result;
    }
};  

2.permutations 全排列(46)

给定一个没有重复数字的序列，返回其所有可能的全排列。

class Solution {
public:
    vector<vector<int> > permute(vector<int> &num) {
	    vector<vector<int> > result;
	    
	    permuteRecursive(num, 0, result);
	    return result;
    }
    
    // permute num[begin..end]
    // invariant: num[0..begin-1] have been fixed/permuted
	void permuteRecursive(vector<int> &num, int begin, vector<vector<int> > &result)	{
		if (begin >= num.size()) {
		    // one permutation instance
		    result.push_back(num);
		    return;
		}
		
		for (int i = begin; i < num.size(); i++) {
		    swap(num[begin], num[i]);
		    permuteRecursive(num, begin + 1, result);
		    // reset
		    swap(num[begin], num[i]);
		}
    }
};

参考链接： https://leetcode.com/problems/permutations/discuss/?currentPage=1&orderBy=most_votes&query=