代码随想录算法训练营第二十二天 | 77. 组合 216.组合总和III 17.电话号码的字母组合-优快云博客

77. 组合

基础题，套用回溯模板直接可以解出来。此处注意要剪枝

class Solution {
public:
    vector<int> res;
    vector<vector<int>> ans; 
    int startID = 1;
void traversal(int startID, int n, int k) {    
        if (res.size() == k) {
            ans.push_back(res);
            return;
        }
        for(int i = startID; i <= n - (k - res.size()) + 1; ++i) {
            res.push_back(i);
            traversal(i+1,n,k);
            res.pop_back();
        }    

    }
    vector<vector<int>> combine(int n, int k) {
        traversal(startID,n,k);
        return ans;
    }
};

216.组合总和III

可以画一张示意图，这里要多保存sum参数用来比较大小，条件都放到最上面，重点是剪枝，不然会超时。

class Solution {
public:
    vector<int> path;
    vector<vector<int>> res;
    void traversal(int startID, int k, int n, int sum) {
        if(sum > n) {
            return;
        }        
        if(path.size() == k) {
            if(sum == n){
                res.push_back(path);
                return;                
            }
        }
        for(int i = startID; i <= 9 - (k - path.size()) + 1; ++i) {
            sum += i;
            path.push_back(i);
            traversal(i+1, k, n, sum);
            path.pop_back(); 
            sum -= i; 
        }
    }
    vector<vector<int>> combinationSum3(int k, int n) {
        traversal(1,k,n,0);
        return res;
    }
};

17.电话号码的字母组合

综合性很强，采用数字映射字符串的方式表示回溯。根据模板图可得此处横向遍历数字对应的字符串，而纵向则是遍历一连串数。

class Solution {
public:
    array<string,10> arr{"","","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"};
    string path;
    vector<string> ans;
    void traversal(string digits, int index) {
        if(digits.empty()) {
            return;
        }
        if(path.size() == digits.size()) {
            ans.push_back(path);
            return; 
        }
        //数字转化成字符串
        int num = (digits[index]) - '0';
        auto str = arr[num]; 
        for(int i = 0; i < str.size(); ++i) {
            auto s = str[i];
            path.push_back(s);
            traversal(digits,index + 1);
            path.pop_back();
        }
    }
    vector<string> letterCombinations(string digits) {
        traversal(digits,0);
        return ans;
    }
};