package LeetCode;

import java.util.ArrayList;
import java.util.LinkedList;
import java.util.List;

public class LeetCode77 {

    //https://leetcode-cn.com/problems/combinations/description/

    /// 时间复杂度: O(n^k)
    /// 空间复杂度: O(k)

    private ArrayList<List<Integer>> res;

    public List<List<Integer>> combine(int n, int k) {

        res = new ArrayList<>();
        if(n <= 0 || k <= 0 || k > n)
            return res;

        LinkedList<Integer> c = new LinkedList<>();
        generateCombinations(n, k, 1, c);

        return res;
    }

    // 求解C(n,k), 当前已经找到的组合存储在c中, 需要从start开始搜索新的元素
    private void generateCombinations(int n, int k, int start, LinkedList<Integer> c){

        if(c.size() == k){
            res.add((List<Integer>)c.clone());
            return;
        }
        for(int i = start ; i <= n ; i ++){
            c.addLast(i);
            generateCombinations(n, k, i + 1, c);
            c.removeLast();
        }

        return;
    }


    // 求解C(n,k), 当前已经找到的组合存储在c中, 需要从start开始搜索新的元素
    private void generateCombinations2(int n, int k, int start, LinkedList<Integer> c){

        if(c.size() == k){
            res.add((List<Integer>)c.clone());
            return;
        }

        // 还有k - c.size()个空位, 所以, [i...n] 中至少要有 k - c.size() 个元素
        // i最多为 n - (k - c.size()) + 1
        // 这种优化叫做截枝
        for(int i = start ; i <= n - (k - c.size()) + 1 ; i ++){
            c.addLast(i);
            generateCombinations2(n, k, i + 1, c);
            c.removeLast();
        }

        return;
    }


    private static void printList(List<Integer> list){
        for(Integer e: list)
            System.out.print(e + " ");
        System.out.println();
    }

    public static void main(String[] args) {

        List<List<Integer>> res = (new LeetCode77()).combine(4, 2);
        for(List<Integer> list: res)
            printList(list);
    }
}