LeetCode 77. Combinations--回溯法，-Python，Java解法

题目地址：

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

Example:

Input: n = 4, k = 2
Output:
[
  [2,4],
  [3,4],
  [2,3],
  [1,2],
  [1,3],
  [1,4],
]

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这道题可以用回溯法来做。
Python回溯解法：

class Solution:
    def combine(self, n: int, k: int) -> List[List[int]]:
        def backtrack(first = 1, curr = []):
            # if the combination is done
            if len(curr) == k:  
                output.append(curr[:])
            for i in range(first, n + 1):
                # add i into the current combination
                curr.append(i)
                # use next integers to complete the combination
                backtrack(i + 1, curr)
                # backtrack
                curr.pop()
        
        output = []
        backtrack()
        return output

第二种解法是使用词典的组合，Python代码如下：

class Solution:
    def combine(self, n: int, k: int) -> List[List[int]]:
        # init first combination
        nums = list(range(1, k + 1)) + [n + 1]
        
        output, j = [], 0
        while j < k:
            # add current combination
            output.append(nums[:k])
            # increase first nums[j] by one
            # if nums[j] + 1 != nums[j + 1]
            j = 0
            while j < k and nums[j + 1] == nums[j] + 1:
                nums[j] = j + 1
                j += 1
            nums[j] += 1
            
        return output

java 回溯解法：

class Solution {
  List<List<Integer>> output = new LinkedList();
  int n;
  int k;

  public void backtrack(int first, LinkedList<Integer> curr) {
    // if the combination is done
    if (curr.size() == k)
      output.add(new LinkedList(curr));

    for (int i = first; i < n + 1; ++i) {
      // add i into the current combination
      curr.add(i);
      // use next integers to complete the combination
      backtrack(i + 1, curr);
      // backtrack
      curr.removeLast();
    }
  }

  public List<List<Integer>> combine(int n, int k) {
    this.n = n;
    this.k = k;
    backtrack(1, new LinkedList<Integer>());
    return output;
  }
}

java 的改进解法：

class Solution {
  public List<List<Integer>> combine(int n, int k) {
    // init first combination
    LinkedList<Integer> nums = new LinkedList<Integer>();
    for(int i = 1; i < k + 1; ++i)
      nums.add(i);
    nums.add(n + 1);

    List<List<Integer>> output = new ArrayList<List<Integer>>();
    int j = 0;
    while (j < k) {
      // add current combination
      output.add(new LinkedList(nums.subList(0, k)));
      // increase first nums[j] by one
      // if nums[j] + 1 != nums[j + 1]
      j = 0;
      while ((j < k) && (nums.get(j + 1) == nums.get(j) + 1))
        nums.set(j, j++ + 1);
      nums.set(j, nums.get(j) + 1);
    }
    return output;
  }
}