LeetCode--Combinations（动态规划与深度优先搜索）

Combinations

Given two integers n and k, return all possible combinations of k numbers out of 1 ... n.

For example,
If n = 4 and k = 2, a solution is:

[
  [2,4],
  [3,4],
  [2,3],
  [1,2],
  [1,3],
  [1,4],
]

及从1到n中选出k个数，得到所有的结果。题目中没有刻意强调升序。

开始想到的是动态规划，后一个数依赖前面的数字。具体实现如下。

public class Solution {
    public List> combine(int n, int k) {
        if (k > n)
            return null;
        else {
            List> result = new ArrayList>();
            List> resultSwitch = new ArrayList>();
            for (int i1 = 0; i1 < k; i1++) {
                if (i1 == 0) {
                    for (int i2 = 1; i2 <= n - k + 1; i2++) {
                        List temp = new ArrayList();
                        temp.add(i2);
                        result.add(temp);
                    }
                } else {
                    while (result.size() != 0) {
                        List temp = result.get(0);
                        for (int i = temp.get(temp.size() - 1) + 1; i <= n - k + temp.size() + 1; i++) {
                            List tempClone = this.clone(temp);
                            tempClone.add(i);
                            resultSwitch.add(tempClone);
                        }
                        result.remove(0);
                    }
                    List> temp = result;
                    result = resultSwitch;
                    resultSwitch = temp;
                }
            }
            return result;
        }
    }

    private List clone(List source) {
        List result = new ArrayList();
        for (int i = 0; i < source.size(); i++) {
            result.add(source.get(i));
        }
        return result;
    }

}

后来在网上看到dfs深度优先搜寻，代码更简洁。

public class Solution {
    public List> combine(int n, int k) {
        if (k > n)
            return null;
        else {
            List> result = new ArrayList>();
            List rs = new ArrayList();
            this.dfs(n, k, rs, result);
            return result;
        }
    }

    private void dfs(int n, int k, List rs, List> result) {
        if (k == 0) {
            result.add(rs);
            return;
        } else {
            for (int i = rs.size() == 0 ? 1 : rs.get(rs.size() - 1) + 1; i <= n - k + 1; i++) {
                List tempList = this.clone(rs);
                tempList.add(i);
                this.dfs(n, k - 1, tempList, result);
            }
        }
    }

    private List clone(List source) {
        List result = new ArrayList();
        for (int i = 0; i < source.size(); i++) {
            result.add(source.get(i));
        }
        return result;
    }

}