leetcode回溯专题

22.括号生成

class Solution
{
public:
    vector<string> res;
    vector<string> generateParenthesis(int n)
    {
        if (n == 0)
            return {};
        string track;
        //可用的左括号和右括号数量
        backtrack(n, n, track, res);
        return res;
    }
    void backtrack(int left, int right, string &track, vector<string> &res)
    {
        //左括号剩的多，不合理
        if (right < left)
            return;
        //数量小于0，肯定不合法
        if (left < 0 || right < 0)
            return;
        //左右括号正好用完，得到一个合法组合
        if(left==0&&right==0)
        {
            res.push_back(track);
            return;
        }

        //放一个左括号
        track.push_back('(');
        backtrack(left - 1, right, track, res);
        track.pop_back();//撤销选择
        //放一个右括号
        track.push_back(')');
        backtrack(left, right-1, track, res);
        track.pop_back();//撤销选择
    }
};

46.全排列

法一：更好

class Solution
{
public:
	vector<vector<int>> permute(vector<int>& nums)
	{
		vector<vector<int>> result;
		permute(nums, result, 0);
		return result;
	}
	void permute(vector<int>& nums, vector<vector<int>>& result, int first)
	{
		if (first == nums.size())
		{
			result.push_back(nums);
			return;
		}
		for (int i = first; i < nums.size(); i++)
		{
			if (first != i)
				swap(nums[first], nums[i]); //交换两数
			permute(nums, result, first + 1);
			if (first != i)
				swap(nums[first], nums[i]);//换回来
		}
	}
};

法二

class Solution {
public:
    vector<vector<int>> result;  
    vector<vector<int>> permute(vector<int>& nums) {
        vector<int> track;
        vector<bool> flag(nums.size(), false);//标记是否访问过了
        backtrack(nums, track, flag);
        return result;
    }
    void backtrack(vector<int>& nums, vector<int> &track,vector<bool>&flag) {
        if(track.size() == nums.size()) {
            result.push_back(track);
            return;
        }
        int nums_size = nums.size();
        for(int i = 0; i < nums_size; ++ i) {
            //如果nums[i]还没被选择，即可供选择
            if(!flag[i]) {
                track.push_back(nums[i]);
                flag[i] = true;
                backtrack(nums, track,flag);
                track.pop_back();
                flag[i] = false;
            }
        }
    }
};

47.全排列II

class Solution
{
public:
	vector<vector<int>> result;
	vector<vector<int>> permuteUnique(vector<int>& nums)
	{
		vector<int> track;
		sort(nums.begin(), nums.end());//排序
		vector<bool> flag(nums.size(), false); //标记是否反问过了
		backtrack(nums, track, flag);
		return result;
	}
	void backtrack(vector<int> nums, vector<int> track, vector<bool> flag)
	{
		if (track.size() == nums.size()) 
		{
			result.push_back(track);
			return;
		}
		int nums_size = nums.size();
		for (int i = 0; i < nums_size; ++i)
		{
			//如果nums[i]没有出现在track中，即可供选择
			if (!flag[i])
			{
				//和上一题比，就增加了排序+下面两句
				//当与前面的数字重复，并且前面的这数字被撤销了而不是正在使用
				if (i > 0 && nums[i] == nums[i - 1] && !flag[i - 1])//剪枝，避免重复
					continue;
				track.push_back(nums[i]);
				flag[i] = true;
				backtrack(nums, track, flag);
				track.pop_back();
				flag[i] = false;
			}
		}
	}
};

78.子集

class Solution {
public:
	vector<vector<int>> res;
    vector<vector<int>> subsets(vector<int>& nums) {
        // 记录走过的路径
        vector<int> track;
        backtrack(nums, 0, track);
        return res;
    }

    void backtrack(vector<int>& nums, int start, vector<int>& track) {
        res.push_back(track);
        // 注意 i 从 start 开始递增
        for (int i = start; i < nums.size(); i++) {
            // 做选择
            track.push_back(nums[i]);
            // 回溯
            backtrack(nums, i + 1, track);
            // 撤销选择
            track.pop_back();
        }
    }
};