regular leetcode

Author: KrisYu

077._combinations.md

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+###77. Combinations
+
+
+题目： 
+<https://leetcode.com/problems/combinations/>
+
+
+难度 : Medium
+
+
+思路一：
+python作弊法
+
+```
+import itertools
+p = [4, 8, 15, 16, 23, 42]
+c = itertools.combinations(p, 4)
+for i in c:
+  print i
+ 
+结果：
+
+(4, 8, 15, 16)
+(4, 8, 15, 23)
+(4, 8, 15, 42)
+(4, 8, 16, 23)
+(4, 8, 16, 42)
+(4, 8, 23, 42)
+(4, 15, 16, 23)
+(4, 15, 16, 42)
+(4, 15, 23, 42)
+(4, 16, 23, 42)
+(8, 15, 16, 23)
+(8, 15, 16, 42)
+(8, 15, 23, 42)
+(8, 16, 23, 42)
+(15, 16, 23, 42)
+```
+
+作弊AC代码:
+
+```
+class Solution(object):
+    def combine(self, n, k):
+        """
+        :type n: int
+        :type k: int
+        :rtype: List[List[int]]
+        """
+        import itertools
+        return [list(i) for i in itertools.combinations(range(1,n+1), k)]
+```
+
+
+思路二：
+
+标准的recursion
+
+但是会超时
+
+
+```
+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 k == 0 :
+    		ans.append(lst)
+    		return
+    	for i in range(start, n+1):
+    		self.dfs(n, k - 1, i + 1,lst +[i], ans)
+```
+
+理解方式
+
+```
+
+					1          2     3
+			    12  13 14    23 24   34			
+```
+
+可以参照这里
+
+
+<http://www.geeksforgeeks.org/print-all-possible-combinations-of-r-elements-in-a-given-array-of-size-n/>
+
+
+解法三：
+
+
+采用递归的方式，在n个数中选k个，如果n大于k，那么可以分类讨论，如果选了n，那么就是在1到(n-1)中选(k-1)个，否则就是在1到(n-1)中选k个。递归终止的条件是k为1，这时候1到n都符合要求。
+
+注意一开始这里的else part花了我一点时间来理解，因为n必定大于k，所以这样递归当 n == k的时候选法就是code原作者的写法，也就是直接[range(1,k+1)]
+
+参考这里： <https://shenjie1993.gitbooks.io/leetcode-python/content/077%20Combinations.html>
+
+
+```
+class Solution(object):
+    def combine(self, n, k):
+        """
+        :type n: int
+        :type k: int
+        :rtype: List[List[int]]
+        """
+        if k == 1:
+            return [[i + 1] for i in range(n)]
+        result = []
+        if n > k:
+            result = [r + [n] for r in self.combine(n - 1, k - 1)] + self.combine(n - 1, k)
+        else: #n == k 
+        	# result = [r + [n] for r in self.combine(n - 1, k - 1)]
+            result = [range(1,k+1)]
+        return result
+```