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文章目录

• Leetcode 17-电话号码的字母组合
• • 题目描述
  • 解题思路
• Leetcode 39-组合总数
• • 题目描述
  • 解题思路
• Leetcode 216-组合总数 Ⅲ
• • 题目描述
  • 解题思路

Leetcode 17-电话号码的字母组合

题目描述

https://leetcode.cn/problems/letter-combinations-of-a-phone-number/description/

解题思路

class Solution {
public:
    vector<string> res;
    string path = "";
    vector<string> list = {"abc", "def","ghi","jkl","mno","pqrs","tuv","wxyz"};
    void backTracking(string s, int startIndex){
        if (startIndex == s.size()){
            res.push_back(path);
            return;
        }
        int digits = s[startIndex]-'0'-2;
        for (int i = 0; i < list[digits].size();i++){
            path += list[digits][i];
            backTracking(s, startIndex+1);
            path.pop_back();//注意字符串移除的方式，不能使用-=
        }
    }
    vector<string> letterCombinations(string digits) {
        if (digits.size()==0) return {};
        backTracking(digits,0);
        return res;
    }
};

Leetcode 39-组合总数

题目描述

https://leetcode.cn/problems/combination-sum/description/

解题思路

class Solution {
public:
    vector<vector<int>> res;
    vector<int> path;
    void backTracking(vector<int>& candidates, int target, int startIndex, int sum){
        if (sum == target){
            res.push_back(path);
            return;
        }
        if (sum >target) return;
        for (int i = startIndex; i < candidates.size(); i++){
            path.push_back(candidates[i]);sum+=candidates[i];
            backTracking(candidates, target, i, sum);
            path.pop_back();sum-=candidates[i];
        }
    }
    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        backTracking(candidates,target,0,0);
        return res;
    }
};

Leetcode 216-组合总数 Ⅲ

题目描述

https://leetcode.cn/problems/combination-sum-iii/description/

解题思路

class Solution {
public:
    vector<vector<int>> res;
    vector<int> path;
    void backTracking(int k, int n, int startIndex, int sum){
        if (sum > n) return;//剪枝操作1
        if (path.size() == k){
            if (sum == n){
                res.push_back(path);
            }
            return;
        }
        for (int i = startIndex; 9-i+1+path.size() >= k; i++){//剪枝操作2.保证剩余的元素个数能满足使组合个数符合k的要求
            path.push_back(i); sum+=i;
            backTracking(k, n, i+1, sum);
            path.pop_back();sum-=i;
        }
    }
    vector<vector<int>> combinationSum3(int k, int n) {
        backTracking(k,n, 1, 0);
        return res;
    }
};