代码随想录算法训练营第25天 | 第七章 回溯算法part04

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今日记录

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491.递增子序列

Leetcode链接

class Solution {
private:
    vector<vector<int>> result;
    vector<int> path;
    void backtracking(vector<int>& nums, int index) {
        if (path.size() > 1) {
            result.push_back(path);
        }
        unordered_set<int> uset;
        for (int i = index; i < nums.size(); i++) {
            if (!path.empty() && nums[i] < path.back() ||
                uset.find(nums[i]) != uset.end()) {
                continue;
            }
            uset.insert(nums[i]);
            path.push_back(nums[i]);
            backtracking(nums, i + 1);
            path.pop_back();
        }
    }

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

46.全排列

Leetcode链接

排列问题：[1,2] 和 [2,1]是两个集合——不需要index
但是纵向的排列问题需要一个vector used数组来标记是否使用过这个元素

class Solution {
private:
    vector<vector<int>> result;
    vector<int> path;
    void backtracking(vector<int>& nums, vector<bool>& used) {
        if (path.size() == nums.size()) {
            result.push_back(path);
            return;
        }
        for (int i = 0; i < nums.size(); i++) {
            if (used[i] == true)
                continue;
            used[i] = true;
            path.push_back(nums[i]);
            backtracking(nums, used);
            path.pop_back();
            used[i] = false;
        }
    }

public:
    vector<vector<int>> permute(vector<int>& nums) {
        result.clear();
        path.clear();
        vector<bool> used(nums.size(), false);
        backtracking(nums, used);
        return result;
    }
};

47.全排列Ⅱ

Leetcode链接

class Solution {
private:
    vector<vector<int>> result;
    vector<int> path;
    void backtracking(vector<int>& nums, vector<bool> used) {
        if (path.size() == nums.size()) {
            result.push_back(path);
            return;
        }
        for (int i = 0; i < nums.size(); i++) {
            // 理解used[i]的含义：used[i-1]==false && nums[i]==nums[i-1]时表明同一层树层两个相同的数，跳过
            // used[i-1]==true && nums[i-1]==nums[i]表明同一树枝有相同的元素，可以重复选取
            if (i > 0 && used[i - 1] == false && nums[i] == nums[i - 1])
                continue;
            if (used[i] == false) {
                used[i] = true;
                path.push_back(nums[i]);
                backtracking(nums, used);
                path.pop_back();
                used[i] = false;
            }
        }
    }

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

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总结