代码随想录算法训练营第二十七天 | 39.组合总和

39.组合总和

题目链接 文章讲解 视频讲解

class Solution {

private:
    vector<int> combine;
    vector<vector<int>> result;
    int count = 0;
public:
    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        backtracking(candidates, target, 0);
        return result;
    }
 
    void backtracking(vector<int>& candidates, int target, int startIndex) {
        // 终止条件 如果count已经大于target，那么就不能继续累加了，立即返回，并执行回溯
        if(count > target) return;
        // 回收结果
        if(count == target) {
            result.push_back(combine);
            return ;
        }

        for(int i = startIndex; i < candidates.size(); ++i) {
            
            combine.push_back(candidates[i]);
            count += candidates[i];
            // startIndex = i表示可以重复选取元素，i + 1表示不重复选取
            backtracking(candidates, target, i);
            count -= candidates[i];
            
            combine.pop_back();
        }
    }
};

40.组合总和II

题目链接 文章讲解 视频讲解
两种去重：

• 树枝去重：再for循环中去重，集合间去重对应的是树层去重
• 树层去重：再递归中去重，集合中去重对应的是树层去重

本题集合中允许重复元素，集合间不允许重复，所以是树枝去重，去重逻辑应写在for循环中

class Solution {
private:
    vector<int> combine;
    vector<vector<int>> result;
    int sum;
public:
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        sort(candidates.begin(), candidates.end());
        backtracking(candidates, target, 0);
        return result;
    }
    
    void backtracking(vector<int>& candidates, int target, int startIndex) {
        // 终止条件
        if(sum > target) return ;
        // 收集结果
        if(sum == target) {
            result.push_back(combine);
            return ;
        }

        for(int i = startIndex; i < candidates.size(); ++i) {
            // 树枝间去重，在每一次选树枝的时候保证和前一个树枝不同
            if(i > startIndex && candidates[i] == candidates[i - 1]) continue;
            combine.push_back(candidates[i]);
            sum += candidates[i];
            backtracking(candidates, target, i + 1);
            sum -= candidates[i];
            combine.pop_back();
        }
    }
};

131.分割回文串

题目链接 文章讲解 视频讲解

class Solution {
private:
    vector<string> split;
    vector<vector<string>> result;
public:
    vector<vector<string>> partition(string s) {
        backtracking(s, 0);
        return result;
    }

    bool isPalindromic(string s) {
        int left = 0, right = s.size() - 1;
        while(left < right) if(s[left++] != s[right--]) return false;
        return true;
    }

    void backtracking(string& s, int startIndex) {
        // 终止条件
        if(startIndex == s.size()) {
            result.push_back(split);
            return ;
        }

        for(int i = startIndex; i < s.size(); ++i) {
            // 判断是否是回文串
            string str = s.substr(startIndex, i - startIndex + 1);
            if(isPalindromic(str)) 
                split.push_back(str);
            else continue;
            backtracking(s, i + 1);
            split.pop_back();
        }
    }
};