Day.22 | 491.递增子序列 46.全排列 47.全排列 II

491.递增子序列

要点：在不排序的前提下，如何去重

class Solution {
public:
    vector<vector<int>> result;
    vector<int> path;
    void backtracking(vector<int>& nums, int startIndex) {
        if (path.size() > 1) {
            result.push_back(path);
        }
        unordered_set<int> uset;
        for (int i = startIndex; 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();
        }
    }

    vector<vector<int>> findSubsequences(vector<int>& nums) {
        backtracking(nums, 0);
        return result;
    }
};

46.全排列

要点：排列问题 与 组合、子集问题的区别。没有startIndex，全部从0开始遍历，不同顺序的排列被认为是不同的，所以需要记录元素是否被使用过，以便同样的元素组合不同的顺序

class Solution {
public:
    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;
        }
    }

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

47.全排列 II

要点：排列问题的去重

class Solution {
public:
    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 (i > 0 && nums[i] == nums[i - 1] && used[i - 1] == false) {
                continue;
            }
            if (used[i] == false) {
                used[i] = true;
                path.push_back(nums[i]);
                backtracking(nums, used);
                path.pop_back();
                used[i] = false;
            }
        }
    }

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