LeetCode----Combinations

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],]

分析：

组合问题。适合使用回溯法解决，另外python下地itertools中也提供了combinations函数。

回溯法：

class Solution(object):    def combine(self, n, k):        """        :type n: int        :type k: int        :rtype: List[List[int]]        """        ans = []        self.dfs(n, k, 1, [], ans)        return ans    def dfs(self, n, k, start, lst, ans):        if not k:            ans.append(lst)            return        for i in range(start, n + 1):            self.dfs(n, k - 1, i + 1, lst + [i], ans)

itertools.combinations:

class Solution(object):    def combine(self, n, k):        """        :type n: int        :type k: int        :rtype: List[List[int]]        """        from itertools import combinations        return [list(c) for c in combinations(range(1, n+1), k)]

回溯法2：

class Solution(object):    def combine(self, n, k):        """        :type n: int        :type k: int        :rtype: List[List[int]]        """        ans = []        stack = []        x = 1        while True:            l = len(stack)            print stack, x, 2 + l            if l == k:                ans.append(stack[:])            if l == k or x > n - k + l + 1:                if not stack:                    return ans                x = stack.pop() + 1            else:                stack.append(x)                x += 1

回溯法2来自于LeetCode中的Discuss。

代码解释：Combinations is typical application for backtracking. Two conditions for back track: (1) the stack length is already k (2) the current value is too large for the rest slots to fit in since we are using ascending order to make sure the uniqueness of each combination.

我的翻译：Combinations是典型的适用回溯法的题型。该代码会在满足两种条件下回溯：（1）当栈的长度已经达到k了；（2）当前值太大，无法放入栈中（由于是升序添加元素，所以可以确保每次产生的解不重复）。