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

题目地址：

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],
]

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 doneif len(curr) == k:  output.append(curr[:])for i in range(first, n + 1):# add i into the current combinationcurr.append(i)# use next integers to complete the combinationbacktrack(i + 1, curr)# backtrackcurr.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 combinationnums = list(range(1, k + 1)) + [n + 1]output, j = [], 0while j < k:# add current combinationoutput.append(nums[:k])# increase first nums[j] by one# if nums[j] + 1 != nums[j + 1]j = 0while j < k and nums[j + 1] == nums[j] + 1:nums[j] = j + 1j += 1nums[j] += 1return 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 doneif (curr.size() == k)output.add(new LinkedList(curr));for (int i = first; i < n + 1; ++i) {// add i into the current combinationcurr.add(i);// use next integers to complete the combinationbacktrack(i + 1, curr);// backtrackcurr.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 combinationLinkedList<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 combinationoutput.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;}
}

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