完成 60. Permutation Sequence

aelam · aelam · commit 02e910a74a57 · 2017-05-27T18:06:26.000+08:00
diff --git a/Algorithms/100. Same Tree/main.swift b/Algorithms/100. Same Tree/main.swift
@@ -0,0 +1,65 @@
+//
+//  main.swift
+//  100. Same Tree
+//
+//  Created by ryan on 26/05/2017.
+//  Copyright © 2017 ryan. All rights reserved.
+//
+
+/*
+Given two binary trees, write a function to check if they are equal or not.
+
+Two binary trees are considered equal if they are structurally identical and the nodes have the same value.
+ 
+ */
+
+/**
+ * Definition for a binary tree node.
+ * public class TreeNode {
+ *     public var val: Int
+ *     public var left: TreeNode?
+ *     public var right: TreeNode?
+ *     public init(_ val: Int) {
+ *         self.val = val
+ *         self.left = nil
+ *         self.right = nil
+ *     }
+ * }
+ */
+
+public class TreeNode {
+    public var val: Int
+    public var left: TreeNode?
+    public var right: TreeNode?
+    public init(_ val: Int) {
+        self.val = val
+        self.left = nil
+        self.right = nil
+    }
+}
+
+class Solution {
+    func isSameTree(_ p: TreeNode?, _ q: TreeNode?) -> Bool {
+        if p == nil && q == nil {
+            return true
+        } else if p != nil && q != nil && p!.val == q!.val {
+            return (isSameTree(p?.left, q?.left) &&
+                    isSameTree(p?.right, q?.right))
+        } else {
+            return false
+        }
+    }
+}
+
+
+let solution = Solution()
+
+
+
+
+
+
+
+
+
+
diff --git a/Algorithms/199. Binary Tree Right Side View/main.swift b/Algorithms/199. Binary Tree Right Side View/main.swift
@@ -0,0 +1,125 @@
+//
+//  main.swift
+//  199. Binary Tree Right Side View
+//
+//  Created by ryan on 26/05/2017.
+//  Copyright © 2017 ryan. All rights reserved.
+//
+
+/*
+Given a binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can see ordered from top to bottom.
+
+For example:
+Given the following binary tree,
+          1            <---
+        /   \
+       2     3         <---
+        \     \
+         5     4       <---
+
+ You should return [1, 3, 4].
+*/
+
+
+/**
+ Definition for a binary tree node.
+ public class TreeNode {
+     public var val: Int
+     public var left: TreeNode?
+     public var right: TreeNode?
+     public init(_ val: Int) {
+         self.val = val
+         self.left = nil
+         self.right = nil
+     }
+ }
+ */
+
+// 按层遍历
+
+public class TreeNode {
+    public var val: Int
+    public var left: TreeNode?
+    public var right: TreeNode?
+    public init(_ val: Int) {
+        self.val = val
+        self.left = nil
+        self.right = nil
+    }
+    
+}
+
+class SimpleSolution {
+    func rightSideView(_ root: TreeNode?) -> [Int] {
+        var results = [Int]()
+        
+        right(root, depth: 0, results: &results)
+        
+        return results
+    }
+    
+    
+    func right(_ node: TreeNode?, depth: Int, results: inout [Int]) {
+        guard let node = node else {
+            return
+        }
+        
+        if results.count == depth {
+            results.append(node.val)
+        }
+        
+        right(node.right, depth: depth + 1, results: &results)
+        right(node.left, depth: depth + 1, results: &results)
+        
+    }
+}
+
+
+class Solution {
+    func rightSideView(_ root: TreeNode?) -> [Int] {
+        var visibleValues = [Int]()
+        let p = root
+        var queue = [[TreeNode]]()
+        
+        if p != nil {
+            queue.append([p!])
+        } else {
+            return []
+        }
+        
+        while queue.count > 0 {
+            let node = queue.remove(at: 0)
+            if let first = node.first {
+                visibleValues.append(first.val)
+            }
+            var children = [TreeNode]()
+            for t in node {
+                if t.right != nil {
+                    children.append(t.right!)
+                }
+                if t.left != nil {
+                    children.append(t.left!)
+                }
+            }
+            if children.count > 0 {
+                queue.append(children)
+            }
+        }
+        
+        return visibleValues
+    }
+}
+
+
+let solution = SimpleSolution()
+let root = TreeNode(1)
+let p2 = TreeNode(2)
+let p3 = TreeNode(3)
+let p4 = TreeNode(4)
+
+
+root.left = p2
+root.right = p3
+p2.left = p4
+
+print(solution.rightSideView(root))
diff --git a/Algorithms/41. First Missing Positive/main.swift b/Algorithms/41. First Missing Positive/main.swift
@@ -0,0 +1,35 @@
+//
+//  main.swift
+//  41. First Missing Positive
+//
+//  Created by ryan on 25/05/2017.
+//  Copyright © 2017 ryan. All rights reserved.
+//
+
+/*
+Given an unsorted integer array, find the first missing positive integer.
+
+For example,
+Given [1,2,0] return 3,
+and [3,4,-1,1] return 2.
+
+Your algorithm should run in O(n) time and uses constant space.
+
+*/
+
+
+class Solution {
+    func firstMissingPositive(_ nums: [Int]) -> Int {
+        var min = nums.first - 1
+        var max = nums.first + 1
+        
+        for i in 1..<nums.count {
+            
+            
+            
+        }
+        
+        
+    }
+}
+
diff --git a/Algorithms/45. Jump Game II/main.swift b/Algorithms/45. Jump Game II/main.swift
@@ -24,6 +24,46 @@
  
  */
 
+class Solution2 {
+    func jump(_ nums: [Int]) -> Int {
+        let count = nums.count
+        
+        if count <= 1 {
+            return 0
+        }
+        
+        var steps = [Int](repeating: Int.min, count: count)
+        var stepCount = 0
+        steps[stepCount] = nums[0]
+        if steps[stepCount] == 0 {
+            return 0
+        }
+        
+        if steps[stepCount] >= count - 1 {
+            return stepCount + 1
+        }
+        for i in 1..<count {
+            if steps[stepCount] >= i {
+                steps[stepCount+1] = max(steps[stepCount+1], steps[stepCount] + nums[i])
+
+                if steps[stepCount] >= count - 1 {
+                    return stepCount
+                }
+            } else {
+                steps[stepCount+1] = max(steps[stepCount+1], steps[stepCount] + nums[i])
+                if steps[stepCount+1] >= count - 1 {
+                    return stepCount+2
+                }
+                stepCount += 1
+            }
+        }
+        
+        return 1
+
+    }
+    
+}
+
 
 class Solution {
     func jump(_ nums: [Int]) -> Int {
@@ -65,13 +105,14 @@ class Solution {
 
 
 
-let solution = Solution()
+let solution = Solution2()
 
 //          0           5          10         15         20         25         30         35   38
-//var data = [5,6,4,4,6,  9,4,4,7,4, 4,8,2,6,8, 1,5,9,6,5, 2,7,9,7,9, 6,9,4,1,6, 8,8,4,4,2, 0,3,8,5]
-var data = [5,6,4,4,6,  9,4,4,7,4, 4,8,2,6,8, 1,5,9,6,5, 2,7,9,7,9, 6,9,4,1,6, 8,8,4,4,2,3,8,5]
+var data = [5,6,4,4,6,  9,4,4,7,4, 4,8,2,6,8, 1,5,9,6,5, 2,7,9,7,9, 6,9,4,1,6, 8,8,4,4,2, 0,3,8,5]
+//var data = [5,6,4,4,6,  9,4,4,7,4, 4,8,2,6,8, 1,5,9,6,5, 2,7,9,7,9, 6,9,4,1,6, 8,8,4,4,2,3,8,5]
 //var data = [5,6,4,4,6,  9,4,4,7,4, 4,8,2,6,8]
-print(solution.jump(data))
+print(solution.jump([1,2]) == 1)
+print(solution.jump([1,2,3]) == 2)
 //print(data.count)
 //print(solution.sol(&data, 38) == 0)
 //print(solution.sol(&data, 36) == 1)
diff --git a/Algorithms/60. Permutation Sequence/main.swift b/Algorithms/60. Permutation Sequence/main.swift
@@ -0,0 +1,87 @@
+//
+//  main.swift
+//  60. Permutation Sequence
+//
+//  Created by ryan on 27/05/2017.
+//  Copyright © 2017 ryan. All rights reserved.
+//
+
+/*
+ 
+ The set [1,2,3,…,n] contains a total of n! unique permutations.
+ 
+ By listing and labeling all of the permutations in order,
+ We get the following sequence (ie, for n = 3):
+ 
+ "123"
+ "132"
+ "213"
+ "231"
+ "312"
+ "321"
+ Given n and k, return the kth permutation sequence.
+ 
+ Note: Given n will be between 1 and 9 inclusive.
+ 
+ */
+
+
+class Solution {
+    func getPermutation(_ n: Int, _ k: Int) -> String {
+        if n == 1 {
+            return "1"
+        }
+        
+        var intArray = (1...n).map { $0 }
+        let permuCount = A(n - 1, 1)
+        return getPermutation(&intArray, permuCount, k - 1)
+    }
+    
+    func getPermutation(_ input: inout [Int], _ permuCount: Int , _ k: Int) -> String {
+        if input.count <= 1 {
+            return String(input.first!)
+        }
+        
+        var results = String()
+        let outputIndex = k / permuCount
+        let output = input.remove(at: outputIndex)
+        results += String(output)
+
+        let newPermuCount = permuCount / input.count
+        results += getPermutation(&input, newPermuCount, k % permuCount)
+        
+        return results
+        
+    }
+
+    func getbang(_ n: Int) -> Int {
+        let f = [1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880]
+        return f[n]
+    }
+
+    func A(_ n: Int, _ m: Int) -> Int {
+        if m == 1 {
+            return getbang(n)
+        }
+        
+        var result = 1
+        if n == m {
+            return n
+        }
+        for i in m...n {
+            result *= i
+        }
+        return result
+    }
+}
+
+
+
+let solution = Solution()
+//print(solution.A(4, 1))
+print(solution.getPermutation(2, 2))
+print(solution.getPermutation(4, 1))
+print(solution.getPermutation(4, 2))
+print(solution.getPermutation(4, 3))
+print(solution.getPermutation(4, 4))
+print(solution.getPermutation(4, 5))
diff --git a/Algorithms/Algorithms.xcodeproj/project.pbxproj b/Algorithms/Algorithms.xcodeproj/project.pbxproj
