回溯法(2)

原题:

/**
 * Created by pradhang on 3/8/2017.
 * Given two integers n and k, return all possible combinations of k numbers out of 1 ... n.
 * <p>
 * For example,
 * If n = 4 and k = 2, a solution is:
 * <p>
 * [
 * [2,4],
 * [3,4],
 * [2,3],
 * [1,2],
 * [1,3],
 * [1,4],
 * ]
 */

答案:


public class Combinations {

    public static void main(String[] args) throws Exception {
        List<List<Integer>> result = new Combinations().combine(3, 3);
    }

    public List<List<Integer>> combine(int n, int k) {
        int[] subArr = new int[k];
        List<List<Integer>> result = new ArrayList<>();
        getNext(0, 0, n, k, subArr, result);
        return result;
    }

    private void getNext(int i, int count,
                         int n, int k, int[] subArr, List<List<Integer>> result) {
        if (k == 0) {
            List<Integer> subList = new ArrayList<>();
            for (int a : subArr)
                subList.add(a);
            result.add(subList);
        } else {
            for (int j = i + 1; j <= n; j++) {
                subArr[count] = j;
                getNext(j, count + 1, n, k - 1, subArr, result);
            }
        }
    }
}
    原文作者:回溯法
    原文地址: https://blog.csdn.net/u011747152/article/details/78351278
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