day25打卡

216. 组合总和 III

• 画出决策树

• 递归出口：path.size() == k
• 子问题：从pos开始枚举每个数字
• 剪枝：count 提前大于 n，返回即可
• 时间复杂度：O(N * 2^N)，空间复杂度：O(N)

class Solution {
public:vector<vector<int>> ret;vector<int> path;int count = 0;vector<vector<int>> combinationSum3(int k, int n) {dfs(k, n, 1);return ret;}void dfs(int k, int n, int pos){//剪枝if(count > n) return;//出口if(path.size() == k){if(count == n)  ret.push_back(path);return;}//子问题for(int i = pos; i <= 9; i++){path.push_back(i);count += i;dfs(k, n, i+1);//回溯path.pop_back();count -= i;}}
};

17. 电话号码的字母组合

• 画出决策树

• 递归出口：pos等于digits.size()
• 子问题：从digits[0]开始枚举每个字母，直到hash[digits[pos] - ‘0’].szie()

class Solution {
public:vector<string> ret;string path = "";string hash[10] = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};vector<string> letterCombinations(string digits) {if(digits.size() == 0) return ret;dfs(digits, 0);return ret;}void dfs(string& digits, int pos){//出口if(pos == digits.size()){ret.push_back(path);return;}//子问题for(int i = 0; i < hash[digits[pos] - '0'].size(); i++){path += hash[digits[pos] - '0'][i];dfs(digits, pos + 1);//回溯path.pop_back();}}
};

pos + 1);
//回溯
path.pop_back();
}
}
};