算法训练day29 LeetCode 39.40.131

39.组合总和

题目

39. 组合总和 - 力扣（LeetCode）

题解

代码随想录 (programmercarl.com)

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

public:
    vector<vector<int>> combinationSum(vector<int> &candidates, int target)
    {
        result.clear();
        path.clear();
        backtracking(candidates, target, 0, 0);
        return result;
    }
};

在sum>target时，依然会进入下一层循环，所以可以添加一个剪枝

    void backtracking(vector<int> &candidates, int target, int sum, int startIndex)
    {
       // if (sum > target)
       //     return;
        if (sum == target)
        {
            result.push_back(path);
            return;
        }
        for (int i = startIndex; i < candidates.size() && sum + candidates[i] <= target; i++) //剪枝
        {
            sum += candidates[i];
            path.push_back(candidates[i]);
            backtracking(candidates, target, sum, i);
            sum -= candidates[i];
            path.pop_back();
        }
    }

40.组合总和II

题目

40. 组合总和 II - 力扣（LeetCode）

题解

代码随想录 (programmercarl.com)

candidates 中元素重复、单个结果中元素可以重复、总结果集中不能有结果重复
需要去重 ---> 将组合的形式以树形结构组织
- 单个结果元素可重复 ---> 单次组合（单枝）中不需要去重
- 多个结果之间不能重复 ---> 不同次组合（同一层）中需要去重 ---> 将元素排序之后组合实现去重

class Solution
{
private:
    vector<vector<int>> result;
    vector<int> path;
    void backtracking(vector<int> &candidates, int target, int sum, int startIndex, vector<bool> used)
    {
        if (sum == target)
        {
            result.push_back(path);
            return;
        }
        for (int i = startIndex; i < candidates.size() && sum + candidates[i] <= target; i++)
        {
            if (i > 0 && candidates[i] == candidates[i - 1] && used[i - 1] == false) //同层去重
                continue;
            sum += candidates[i];
            path.push_back(candidates[i]);
            used[i] = true;
            backtracking(candidates, target, sum, i + 1, used);
            used[i] = false;
            sum -= candidates[i];
            path.pop_back();
        }
    }

public:
    vector<vector<int>> combinationSum2(vector<int> &candidates, int target)
    {
        vector<bool> used(candidates.size(), false);
        result.clear();
        path.clear();
        sort(candidates.begin(), candidates.end());
        backtracking(candidates, target, 0, 0, used);
        return result;
    }
};

131.分割回文串

题目

131. 分割回文串 - 力扣（LeetCode）

题解

代码随想录 (programmercarl.com)

递归用来纵向遍历，for循环用来横向遍历，切割线（就是图中的红线）切割到字符串的结尾位置，说明找到了一个切割方法。

class Solution
{
private:
    vector<vector<string>> result;
    vector<string> path;
    bool isPalindrome(const string &s, int start, int end)
    {
        for (int i = start, j = end; i < j; i++, j--)
        {
            if (s[i] != s[j])
                return false;
        }
        return true;
    }
    void backtracking(const string &s, int startIndex)
    {
        if (startIndex >= s.size())
        {
            result.push_back(path);
            return;
        }
        for (int i = startIndex; i < s.size(); i++)
        {
            if (isPalindrome(s, startIndex, i))
            {
                string str = s.substr(startIndex, i - startIndex + 1);
                path.push_back(str);
            }
            else
                continue;
            backtracking(s, i + 1);
            path.pop_back();
        }
    }

public:
    vector<vector<string>> partition(string s)
    {
        result.clear();
        path.clear();
        backtracking(s, 0);
        return result;
    }
};