常见算法
2023-01-09 本文已影响0人
小白lf
class ListNode {
var val: Int
var next: ListNode?
init() { self.val = 0; self.next = nil; }
init(_ val: Int) { self.val = val; self.next = nil; }
init(_ val: Int, _ next: ListNode?) { self.val = val; self.next = next; }
}
class TreeNode {
var val: Int
var left: TreeNode?
var right: TreeNode?
init() { self.val = 0; self.left = nil; self.right = nil; }
init(_ val: Int) { self.val = val; self.left = nil; self.right = nil; }
init(_ val: Int, _ left: TreeNode?, _ right: TreeNode?) {
self.val = val
self.left = left
self.right = right
}
}
//两数之和
func twoSum(_ nums: [Int], _ target: Int) -> [Int] {
if nums.isEmpty { return [] }
var dict = [Int: Int]()
for (index, num) in nums.enumerated() {
if let _index = dict[target-num] {
return [_index, index]
}
dict[num] = index
}
return []
}
//两数相加
func addTwoNumbers(_ l1: ListNode?, _ l2: ListNode?) -> ListNode? {
var l1 = l1, l2 = l2
if l1 == nil && l2 == nil { return nil }
var head: ListNode?
var cur = head
var c = 0
while l1 != nil || l2 != nil {
let a = l1?.val ?? 0
let b = l2?.val ?? 0
let val = (a + b + c) % 10
c = (a + b + c) / 10
if head == nil {
head = ListNode(val)
cur = head
} else {
cur?.next = ListNode(val)
cur = cur?.next
}
l1 = l1?.next
l2 = l2?.next
}
if c == 1 {
cur?.next = ListNode(1)
}
return head
}
//字符串反转
func reverse(_ str: String?) -> String? {
guard let str = str, !str.isEmpty else { return nil }
var strs = Array(str)
var i = 0, j = strs.count - 1
while i < j {
strs.swapAt(i, j)
i += 1
j -= 1
}
return String(strs)
}
//输入: s = "abcabcbb"
//输出: 3
func lengthOfLongestSubstring(_ s: String) -> Int {
var maxLength = 0
var i = 0
var dict = [String: Int]()
for (j, c) in s.enumerated() {
if let index = dict[String(c)] {
i = max(i, index + 1)
}
maxLength = max(maxLength, j - i + 1)
dict[String(c)] = j
}
return maxLength
}
//有效的括号
func isValid(_ s: String) -> Bool {
if s.count % 2 != 0 { return false }
let maps = [")": "(", "]": "[", "}": "{"]
var stacks: [String] = []
for c in s {
let ss = String(c)
if let ss_ = maps[ss] {
if stacks.isEmpty || stacks.last != ss_ {
return false
}
stacks.removeLast()
} else {
stacks.append(ss)
}
}
return stacks.isEmpty
}
//合并两个有序链表
func mergeTwoLists(_ list1: ListNode?, _ list2: ListNode?) -> ListNode? {
guard let list1 = list1 else { return list2 }
guard let list2 = list2 else { return list1 }
if list1.val < list2.val {
list1.next = mergeTwoLists(list1.next, list2)
return list1
} else {
list2.next = mergeTwoLists(list2.next, list1)
return list2
}
}
//爬楼梯
func climbStairs(_ n: Int) -> Int {
var p = 0, q = 0, r = 1
var j = 1
while j <= n {
p = q
q = r
r = p + q
j += 1
}
return r
}
//二叉树的中序遍历
func inorderTraversal(_ root: TreeNode?) -> [Int] {
guard let root = root else { return [] }
var notes: [Int] = []
if root.left != nil {
notes.append(contentsOf: inorderTraversal(root.left))
}
notes.append(root.val)
if root.right != nil {
notes.append(contentsOf: inorderTraversal(root.right))
}
return notes
}
//对称二叉树
func isSymmetric(_ root: TreeNode?) -> Bool {
func check(left: TreeNode?, right: TreeNode?) -> Bool {
if left == nil && right == nil { return true }
if left == nil || right == nil { return false }
return left?.val == right?.val &&
check(left: left?.left, right: right?.right) &&
check(left: left?.right, right: right?.left)
}
return check(left: root, right: root)
}
//二叉树的最大深度
func maxDepth(_ root: TreeNode?) -> Int {
guard let root = root else { return 0 }
return 1 + max(maxDepth(root.left), maxDepth(root.right))
}
//买卖股票的最佳时机
func maxProfit(_ prices: [Int]) -> Int {
var minPrice = Int.max
var maxProfit = 0
for price in prices {
if price < minPrice {
minPrice = price
}
maxProfit = max(maxProfit, price-minPrice)
}
return maxProfit
}
//只出现一次的数字
//给你一个 非空 整数数组 nums ,除了某个元素只出现一次以外,其余每个元素均出现两次。找出那个只出现了一次的元素
func singleNumber(_ nums: [Int]) -> Int {
var dict: [Int: Int] = [:]
for num in nums {
if let res = dict[num] {
dict[num] = res + 1
} else {
dict[num] = 1
}
}
return dict.sorted { $0.value < $1.value }.first?.key ?? 0
// var res = 0
// for num in nums {
// res ^= num
// }
// return res
}
//链表是否有环
func hasCycle(_ head: ListNode?) -> Bool {
if head == nil || head?.next == nil {
return false
}
var slow = head, fast = head?.next
while slow !== fast {
if fast == nil || fast?.next == nil {
return false
}
slow = slow?.next
fast = fast?.next?.next
}
return true
}
//相交链表
func getIntersectionNode(_ headA: ListNode?, _ headB: ListNode?) -> ListNode? {
var pA = headA, pB = headB
if pA == nil || pB == nil { return nil }
while pA !== pB {
pA = pA == nil ? headB : pA?.next
pB = pB == nil ? headA : pB?.next
}
return pA
}
//反转链表
func reverseList(_ head: ListNode?) -> ListNode? {
if head == nil || head?.next == nil { return head }
var head = head
var p1: ListNode? = nil
var p2 = head
while head != nil {
p2 = p2?.next
head?.next = p1
p1 = head
head = p2
}
return p1
}
//翻转二叉树
func invertTree(_ root: TreeNode?) -> TreeNode? {
if root == nil { return nil }
let right = root?.right
let left = root?.left
root?.left = invertTree(right)
root?.right = invertTree(left)
return root
}
//合并二叉树
func mergeTrees(_ root1: TreeNode?, _ root2: TreeNode?) -> TreeNode? {
guard let root1 = root1 else { return root2 }
guard let root2 = root2 else { return root1 }
let root = TreeNode(root1.val + root2.val)
root.left = mergeTrees(root1.left, root2.left)
root.right = mergeTrees(root1.right, root2.right)
return root
}
//二叉树的直径
//给定一棵二叉树,你需要计算它的直径长度。一棵二叉树的直径长度是任意两个结点路径长度中的最大值。这条路径可能穿过也可能不穿过根结点
//思路: 把每一个结点当成根节点 得到其 左子树深度+右子树深度+1, 然后取这些值的最大值 就是二叉树的直径
func diameterOfBinaryTree(_ root: TreeNode?) -> Int {
var res = 0
func maxDepth(_ root: TreeNode?) -> Int {
guard let root = root else { return 0 }
return 1 + max(maxDepth(root.left), maxDepth(root.right))
}
func maxNotes(_ root: TreeNode?) {
guard let root = root else { return }
let l = maxDepth(root.left)
let r = maxDepth(root.right)
res = max(res, l + r)
maxNotes(root.left)
maxNotes(root.right)
}
maxNotes(root)
return res
}
//删除链表的倒数第 N 个结点
func removeNthFromEnd(_ head: ListNode?, _ n: Int) -> ListNode? {
if n <= 0 { return head }
let preHead: ListNode? = ListNode(0, head)
var p = preHead, q = head
for _ in 0..<n {
q = q?.next
}
while q != nil {
p = p?.next
q = q?.next
}
p?.next = p?.next?.next
return preHead?.next
}
//二叉树展开为链表(按照前序遍历顺序)
func flatten(_ root: TreeNode?) {
if root == nil { return }
var p = root
if p?.left != nil {
p = p?.left
while p?.right != nil {
p = p?.right
}
p?.right = root?.right
root?.right = root?.left
root?.left = nil
}
flatten(root?.right)
}
//二叉树的层序遍历
func levelOrder(_ root: TreeNode?) -> [[Int]] {
if root == nil { return [] }
var res: [[Int]] = []
var arr: [TreeNode?] = [root]
while !arr.isEmpty {
let count = arr.count
var temps: [Int] = []
for _ in 0..<count {
let node = arr.removeFirst()
temps.append(node!.val)
if node?.left != nil {
arr.append(node?.left)
}
if node?.right != nil {
arr.append(node?.right)
}
}
res.append(temps)
}
return res
}
//验证二叉搜索树
func isValidBST(_ root: TreeNode?) -> Bool {
func isValidBST(_ note: TreeNode?, min: Int, max: Int) -> Bool {
if note == nil { return true }
let value = note!.val
if value <= min || value >= max {
return false
}
return isValidBST(note?.left, min: min, max: value) &&
isValidBST(note?.right, min: value, max: max)
}
return isValidBST(root, min: Int.min, max: Int.max)
}
// 斐波那契数列
func fib(_ n: Int) -> Int {
if n < 2 { return n }
var p = 0, q = 0, r = 1
for _ in 2...n {
p = q
q = r
r = (p + q) % 1_000_000_007
}
return r
}