backtracking

junncao · junncao · commit c8fd5ed66709 · 2019-10-24T21:39:59.000+08:00
diff --git a/.vscode/launch.json b/.vscode/launch.json
@@ -0,0 +1,70 @@
+{
+    // 使用 IntelliSense 了解相关属性。 
+    // 悬停以查看现有属性的描述。
+    // 欲了解更多信息，请访问: https://go.microsoft.com/fwlink/?linkid=830387
+    "version": "0.2.0",
+    "configurations": [
+        {
+            "name": "Python: Current File (Integrated Terminal)",
+            "type": "python",
+            "request": "launch",
+            "program": "${file}",
+            "console": "integratedTerminal"
+        },
+        {
+            "name": "Python: Remote Attach",
+            "type": "python",
+            "request": "attach",
+            "port": 5678,
+            "host": "localhost",
+            "pathMappings": [
+                {
+                    "localRoot": "${workspaceFolder}",
+                    "remoteRoot": "."
+                }
+            ]
+        },
+        {
+            "name": "Python: Module",
+            "type": "python",
+            "request": "launch",
+            "module": "enter-your-module-name-here",
+            "console": "integratedTerminal"
+        },
+        {
+            "name": "Python: Django",
+            "type": "python",
+            "request": "launch",
+            "program": "${workspaceFolder}/manage.py",
+            "console": "integratedTerminal",
+            "args": [
+                "runserver",
+                "--noreload",
+                "--nothreading"
+            ],
+            "django": true
+        },
+        {
+            "name": "Python: Flask",
+            "type": "python",
+            "request": "launch",
+            "module": "flask",
+            "env": {
+                "FLASK_APP": "app.py"
+            },
+            "args": [
+                "run",
+                "--no-debugger",
+                "--no-reload"
+            ],
+            "jinja": true
+        },
+        {
+            "name": "Python: Current File (External Terminal)",
+            "type": "python",
+            "request": "launch",
+            "program": "${file}",
+            "console": "externalTerminal"
+        }
+    ]
+}
diff --git a/116.填充每个节点的下一个右侧节点指针.py b/116.填充每个节点的下一个右侧节点指针.py
@@ -0,0 +1,81 @@
+#
+# @lc app=leetcode.cn id=116 lang=python
+#
+# [116] 填充每个节点的下一个右侧节点指针
+#
+# https://leetcode-cn.com/problems/populating-next-right-pointers-in-each-node/description/
+#
+# algorithms
+# Medium (48.22%)
+# Likes:    94
+# Dislikes: 0
+# Total Accepted:    15.2K
+# Total Submissions: 31.4K
+# Testcase Example:  '{"$id":"1","left":{"$id":"2","left":{"$id":"3","left":null,"next":null,"right":null,"val":4},"next":null,"right":{"$id":"4","left":null,"next":null,"right":null,"val":5},"val":2},"next":null,"right":{"$id":"5","left":{"$id":"6","left":null,"next":null,"right":null,"val":6},"next":null,"right":{"$id":"7","left":null,"next":null,"right":null,"val":7},"val":3},"val":1}'
+#
+# 给定一个完美二叉树，其所有叶子节点都在同一层，每个父节点都有两个子节点。二叉树定义如下：
+# 
+# struct Node {
+# ⁠ int val;
+# ⁠ Node *left;
+# ⁠ Node *right;
+# ⁠ Node *next;
+# }
+# 
+# 填充它的每个 next 指针，让这个指针指向其下一个右侧节点。如果找不到下一个右侧节点，则将 next 指针设置为 NULL。
+# 
+# 初始状态下，所有 next 指针都被设置为 NULL。
+# 
+# 
+# 
+# 示例：
+# 
+# 
+# 
+# 
+# 输入：{"$id":"1","left":{"$id":"2","left":{"$id":"3","left":null,"next":null,"right":null,"val":4},"next":null,"right":{"$id":"4","left":null,"next":null,"right":null,"val":5},"val":2},"next":null,"right":{"$id":"5","left":{"$id":"6","left":null,"next":null,"right":null,"val":6},"next":null,"right":{"$id":"7","left":null,"next":null,"right":null,"val":7},"val":3},"val":1}
+# 
+# 
+# 输出：{"$id":"1","left":{"$id":"2","left":{"$id":"3","left":null,"next":{"$id":"4","left":null,"next":{"$id":"5","left":null,"next":{"$id":"6","left":null,"next":null,"right":null,"val":7},"right":null,"val":6},"right":null,"val":5},"right":null,"val":4},"next":{"$id":"7","left":{"$ref":"5"},"next":null,"right":{"$ref":"6"},"val":3},"right":{"$ref":"4"},"val":2},"next":null,"right":{"$ref":"7"},"val":1}
+# 
+# 解释：给定二叉树如图 A 所示，你的函数应该填充它的每个 next 指针，以指向其下一个右侧节点，如图 B 所示。
+# 
+# 
+# 
+# 
+# 提示：
+# 
+# 
+# 你只能使用常量级额外空间。
+# 使用递归解题也符合要求，本题中递归程序占用的栈空间不算做额外的空间复杂度。
+# 
+# 
+#
+
+# @lc code=start
+"""
+# Definition for a Node.
+class Node(object):
+    def __init__(self, val, left, right, next):
+        self.val = val
+        self.left = left
+        self.right = right
+        self.next = next
+"""
+class Solution(object):
+    def connect(self, root):
+        """
+        :type root: Node
+        :rtype: Node
+        """
+        if not root or not root.left:return root  #为None 或者 为叶子节点
+        root.left.next=root.right
+        if root.next:
+            root.right.next=root.next.left
+        
+        self.connect(root.left)
+        self.connect(root.right)
+                #要先处理根节点保证同层次之间已经串起来了才可以进行下一层，先左边还是右边的顺序无要求
+        return root
+# @lc code=end
+
diff --git a/51.n皇后.py b/51.n皇后.py
@@ -0,0 +1,81 @@
+#
+# @lc app=leetcode.cn id=51 lang=python
+#
+# [51] N皇后
+#
+# https://leetcode-cn.com/problems/n-queens/description/
+#
+# algorithms
+# Hard (66.02%)
+# Likes:    245
+# Dislikes: 0
+# Total Accepted:    16.6K
+# Total Submissions: 25K
+# Testcase Example:  '4'
+#
+# n 皇后问题研究的是如何将 n 个皇后放置在 n×n 的棋盘上，并且使皇后彼此之间不能相互攻击。
+# 
+# 
+# 
+# 上图为 8 皇后问题的一种解法。
+# 
+# 给定一个整数 n，返回所有不同的 n 皇后问题的解决方案。
+# 
+# 每一种解法包含一个明确的 n 皇后问题的棋子放置方案，该方案中 'Q' 和 '.' 分别代表了皇后和空位。
+# 
+# 示例:
+# 
+# 输入: 4
+# 输出: [
+# ⁠[".Q..",  // 解法 1
+# ⁠ "...Q",
+# ⁠ "Q...",
+# ⁠ "..Q."],
+# 
+# ⁠["..Q.",  // 解法 2
+# ⁠ "Q...",
+# ⁠ "...Q",
+# ⁠ ".Q.."]
+# ]
+# 解释: 4 皇后问题存在两个不同的解法。
+# 
+# 
+#
+
+# @lc code=start
+class Solution(object):
+    def solveNQueens(self, n):
+        """
+        :type n: int
+        :rtype: List[List[str]]
+        """
+        res=[]
+        board=[['.' for i in range(n)] for i in range(n)]
+        col=[True]*n
+        diag1=[True]*(2*n-1)    #左下到右上的许多对角线，从左上到右下从0开始编号
+        diag2=[True]*(2*n-1)    #左上到右下的许多对角线，从左下到右上从0开始编号
+
+        def updateBoard(x,y,n,flag):
+            col[x]=flag
+            diag1[x+y]=flag
+            diag2[x-y+n-1]=flag
+            board[y][x]='Q' if not flag else '.'
+        def available(x,y):
+            return col[x] and diag1[x+y] and diag2[x-y+n-1]
+        def nqueens(y,n):
+            if y>=n:
+                temp=[]
+                for i in range(n):
+                    temp.append(''.join(board[i]))
+                res.append(temp)
+                return
+            for x in range(n):  #同一排挨个试
+                if not available(x,y):
+                    continue
+                updateBoard(x,y,n,False)    #直到出现一个可以填Q的
+                nqueens(y+1,n)  #进行下一行的选取
+                updateBoard(x,y,n,True)     #再调整回来
+        nqueens(0,n)
+        return res 
+# @lc code=end
+
diff --git a/78.子集.py b/78.子集.py
@@ -45,7 +45,7 @@ def subsets(self, nums):
     def dfs(self,index,path,nums,res):
         res.append(path)
         print(path)
-        for i in xrange(index,len(nums)):
+        for i in range(index,len(nums)):
             self.dfs(i+1,path+[nums[i]],nums,res)
 
 #第二种：列表推导式
