add backtracking

Backtracking/216_CombinationSumIII.py

@@ -22,12 +22,12 @@ def _combination_sum(self, k, n, nums):
         # Get the new num from start
         start = 1
         if nums:
-            start = nums[-1]+1
+            start = nums[-1] + 1
         for i in range(start, 10):
             cur_sum = sum(nums) + i
             if cur_sum <= n:
                 nums.append(i)
-                self._combination_sum(k-1, n, nums)
+                self._combination_sum(k - 1, n, nums)
                 del nums[-1]    # Backtracking
             else:
                 break

Backtracking/46_Permutations.py

@@ -13,7 +13,7 @@ def dfs(self, nums, path, ans):
         if not nums:
             ans.append(path)
         for i, n in enumerate(nums):
-            self.dfs(nums[:i]+nums[i+1:], path+[n], ans)
+            self.dfs(nums[:i] + nums[i + 1:], path + [n], ans)
 
 
 class Solution_2(object):
@@ -22,7 +22,7 @@ class Solution_2(object):
     def permute(self, nums):
         return [[n] + p
                 for i, n in enumerate(nums)
-                for p in self.permute(nums[:i] + nums[i+1:])] or [[]]
+                for p in self.permute(nums[:i] + nums[i + 1:])] or [[]]
 
 """
 []

Week07/51.py → Backtracking/51_NQueens.py

@@ -58,7 +58,8 @@ def isValid(self, row, col, matrix, n):
         return True
 
 """
-if __name__ == '__main__':
-    sol = Solution()
-    print sol.solveNQueens(2)
+0
+1
+5
+
 """

Backtracking/79_WordSearch.py

@@ -3,13 +3,7 @@
 
 
 class Solution(object):
-
     def exist(self, board, word):
-        """
-        :type board: List[List[str]]
-        :type word: str
-        :rtype: bool
-        """
         if not board and word:
             return False
         if not word:
@@ -66,13 +60,13 @@ def adj_pos_lists(self, cur_pos, board):
         col = cur_pos[1]
         adj_list = []
         if row - 1 >= 0:
-            adj_list.append([row-1, col])
+            adj_list.append([row - 1, col])
         if row + 1 < m_rows:
-            adj_list.append([row+1, col])
+            adj_list.append([row + 1, col])
         if col - 1 >= 0:
-            adj_list.append([row, col-1])
+            adj_list.append([row, col - 1])
         if col + 1 < n_cols:
-            adj_list.append([row, col+1])
+            adj_list.append([row, col + 1])
         return adj_list
 
 """

Backtracking/93_RestoreIPAddresses.py

@@ -31,7 +31,7 @@ def restoreAddress(self, s, count):
 
         # Current field is '0'
         if s[0] == "0":
-            address_1 = self.restoreAddress(s[1:], count+1)
+            address_1 = self.restoreAddress(s[1:], count + 1)
             for block in address_1:
                 cur_address = ['0']
                 cur_address.extend(block)
@@ -43,7 +43,7 @@ def restoreAddress(self, s, count):
         for i in range(1, 4):
             if len(s) < i or int(s[:i]) > 255:
                 continue
-            address_1 = self.restoreAddress(s[i:], count+1)
+            address_1 = self.restoreAddress(s[i:], count + 1)
             for block in address_1:
                 cur_address = [s[:i]]
                 cur_address.extend(block)

Backtracking/README.md

@@ -49,13 +49,11 @@
 * 如何保存根结点到叶子结点的路径；
 * 如何回退到父结点。
 
-针对不同的问题场景，上面的问题有不同的解决办法，无法一一给出。不过我们可以抽象地给出所有问题的一般解过程，可分为递归实现和迭代实现。
-
-递归实现（保存所有的解）：
+针对不同的问题场景，上面的问题有不同的解决办法，无法一一给出。不过我们可以抽象地给出所有问题的一般解过程，下面为其递归实现（保存所有的解）：
 
     void backtrack(state s) {
         if(s is end){           // 当前结点为可行解
-            sols.append(path)   // 保存该解
+            sols.append(path);  // 保存该解
         }
         else if(s has no ways){ // 当前结点为不可达叶子结点
             return; 
@@ -69,9 +67,6 @@
         }
     }
 
-所有的递归实现均可以转换为迭代实现，只不过会用到额外的空间。回溯法迭代实现（保存所有的解）如下：
-
-
 # 例子：更好的理解