Golang每日一练(leetDay0033)

目录

97. 交错字符串 Interleaving String  

98. 验证二叉搜索树 Validate Binary Search Tree  

99. 恢复二叉搜索树 Recover Binary Search Tree  

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97. 交错字符串 Interleaving String

给定三个字符串 s1s2s3,请你帮忙验证 s3 是否是由 s1 和 s2 交错 组成的。

两个字符串 s 和 t 交错 的定义与过程如下,其中每个字符串都会被分割成若干 非空 子字符串:

  • s = s1 + s2 + ... + sn
  • t = t1 + t2 + ... + tm
  • |n - m| <= 1
  • 交错 是 s1 + t1 + s2 + t2 + s3 + t3 + ... 或者 t1 + s1 + t2 + s2 + t3 + s3 + ...

注意:a + b 意味着字符串 a 和 b 连接。

示例 1:

Golang每日一练(leetDay0033)_第1张图片

输入:s1 = "aabcc", s2 = "dbbca", s3 = "aadbbcbcac"
输出:true

示例 2:

输入:s1 = "aabcc", s2 = "dbbca", s3 = "aadbbbaccc"
输出:false

示例 3:

输入:s1 = "", s2 = "", s3 = ""
输出:true

提示:

  • 0 <= s1.length, s2.length <= 100
  • 0 <= s3.length <= 200
  • s1s2、和 s3 都由小写英文字母组成

进阶:您能否仅使用 O(s2.length) 额外的内存空间来解决它?

代码1:动态规划

package main

import (
	"fmt"
)

func isInterleave(s1 string, s2 string, s3 string) bool {
	if len(s1)+len(s2) != len(s3) {
		return false
	}
	dp := make([][]bool, len(s1)+1)
	for i := range dp {
		dp[i] = make([]bool, len(s2)+1)
	}
	dp[0][0] = true
	for i := 1; i <= len(s1); i++ {
		dp[i][0] = dp[i-1][0] && s1[i-1] == s3[i-1]
	}
	for j := 1; j <= len(s2); j++ {
		dp[0][j] = dp[0][j-1] && s2[j-1] == s3[j-1]
	}
	for i := 1; i <= len(s1); i++ {
		for j := 1; j <= len(s2); j++ {
			if s1[i-1] == s3[i+j-1] {
				dp[i][j] = dp[i][j] || dp[i-1][j]
			}
			if s2[j-1] == s3[i+j-1] {
				dp[i][j] = dp[i][j] || dp[i][j-1]
			}
		}
	}
	return dp[len(s1)][len(s2)]
}

func main() {
	s1 := "aabcc"
	s2 := "dbbca"
	s3 := "aadbbcbcac"
	fmt.Println(isInterleave(s1, s2, s3))
	s1 = "aabcc"
	s2 = "dbbca"
	s3 = "aadbbbaccc"
	fmt.Println(isInterleave(s1, s2, s3))
	s1 = ""
	s2 = ""
	s3 = ""
	fmt.Println(isInterleave(s1, s2, s3))
}

输出:

true
false
true

代码2:广度优先搜索

package main

import (
	"fmt"
)

func isInterleave(s1 string, s2 string, s3 string) bool {
	if len(s1)+len(s2) != len(s3) {
		return false
	}
	queue := [][]int{{0, 0}}
	visited := make([][]bool, len(s1)+1)
	for i := range visited {
		visited[i] = make([]bool, len(s2)+1)
	}
	for len(queue) > 0 {
		i, j := queue[0][0], queue[0][1]
		queue = queue[1:]
		if visited[i][j] {
			continue
		}
		visited[i][j] = true
		if i == len(s1) && j == len(s2) {
			return true
		}
		if i < len(s1) && s1[i] == s3[i+j] {
			queue = append(queue, []int{i + 1, j})
		}
		if j < len(s2) && s2[j] == s3[i+j] {
			queue = append(queue, []int{i, j + 1})
		}
	}
	return false
}

func main() {
	s1 := "aabcc"
	s2 := "dbbca"
	s3 := "aadbbcbcac"
	fmt.Println(isInterleave(s1, s2, s3))
	s1 = "aabcc"
	s2 = "dbbca"
	s3 = "aadbbbaccc"
	fmt.Println(isInterleave(s1, s2, s3))
	s1 = ""
	s2 = ""
	s3 = ""
	fmt.Println(isInterleave(s1, s2, s3))
}

98. 验证二叉搜索树 Validate Binary Search Tree

给你一个二叉树的根节点 root ,判断其是否是一个有效的二叉搜索树。

有效 二叉搜索树定义如下:

  • 节点的左子树只包含 小于 当前节点的数。
  • 节点的右子树只包含 大于 当前节点的数。
  • 所有左子树和右子树自身必须也是二叉搜索树。

示例 1:

Golang每日一练(leetDay0033)_第2张图片

输入:root = [2,1,3]
输出:true

示例 2:

Golang每日一练(leetDay0033)_第3张图片

输入:root = [5,1,4,null,null,3,6]
输出:false
解释:根节点的值是 5 ,但是右子节点的值是 4 。

提示:

  • 树中节点数目范围在[1, 10^4] 内
  • -2^31 <= Node.val <= 2^31 - 1

代码1:

package main

import (
	"fmt"
)

const null = -1 << 31

type TreeNode struct {
	Val   int
	Left  *TreeNode
	Right *TreeNode
}

func isValidBST(root *TreeNode) bool {
	var pre *TreeNode
	return validate(root, &pre)
}

func validate(node *TreeNode, pre **TreeNode) bool {
	if node == nil {
		return true
	}
	if !validate(node.Left, pre) {
		return false
	}
	if *pre != nil && node.Val <= (*pre).Val {
		return false
	}
	*pre = node
	return validate(node.Right, pre)
}

func buildTree(nums []int) *TreeNode {
	if len(nums) == 0 {
		return nil
	}
	root := &TreeNode{Val: nums[0]}
	Queue := []*TreeNode{root}
	idx := 1
	for idx < len(nums) {
		node := Queue[0]
		Queue = Queue[1:]
		if nums[idx] != null {
			node.Left = &TreeNode{Val: nums[idx]}
			Queue = append(Queue, node.Left)
		}
		idx++
		if idx < len(nums) && nums[idx] != null {
			node.Right = &TreeNode{Val: nums[idx]}
			Queue = append(Queue, node.Right)
		}
		idx++
	}
	return root
}

func (root *TreeNode) LevelOrder() []int {
	var res []int
	if root == nil {
		return res
	}
	Queue := []*TreeNode{root}
	for len(Queue) > 0 {
		cur := Queue[0]
		Queue = Queue[1:]
		res = append(res, cur.Val)
		if cur.Left != nil {
			Queue = append(Queue, cur.Left)
		}
		if cur.Right != nil {
			Queue = append(Queue, cur.Right)
		}
	}
	return res
}

func main() {
	nums := []int{2, 1, 3}
	root := buildTree(nums)
	fmt.Println(isValidBST(root))
	nums = []int{5, 1, 4, null, null, 3, 6}
	root = buildTree(nums)
	fmt.Println(isValidBST(root))
	nums = []int{3, 1, 5, null, null, 4, 6}
	root = buildTree(nums)
	fmt.Println(isValidBST(root))
}

输出:

true
false
true

代码2: 递归法

package main

import (
	"fmt"
)

const null = -1 << 31

type TreeNode struct {
	Val   int
	Left  *TreeNode
	Right *TreeNode
}

func isValidBST(root *TreeNode) bool {
	return validate(root, nil, nil)
}

func validate(node *TreeNode, min *int, max *int) bool {
	if node == nil {
		return true
	}
	if (min != nil && node.Val <= *min) || (max != nil && node.Val >= *max) {
		return false
	}
	return validate(node.Left, min, &node.Val) && validate(node.Right, &node.Val, max)
}

func buildTree(nums []int) *TreeNode {
	if len(nums) == 0 {
		return nil
	}
	root := &TreeNode{Val: nums[0]}
	Queue := []*TreeNode{root}
	idx := 1
	for idx < len(nums) {
		node := Queue[0]
		Queue = Queue[1:]
		if nums[idx] != null {
			node.Left = &TreeNode{Val: nums[idx]}
			Queue = append(Queue, node.Left)
		}
		idx++
		if idx < len(nums) && nums[idx] != null {
			node.Right = &TreeNode{Val: nums[idx]}
			Queue = append(Queue, node.Right)
		}
		idx++
	}
	return root
}

func main() {
	nums := []int{2, 1, 3}
	root := buildTree(nums)
	fmt.Println(isValidBST(root))
	nums = []int{5, 1, 4, null, null, 3, 6}
	root = buildTree(nums)
	fmt.Println(isValidBST(root))
	nums = []int{3, 1, 5, null, null, 4, 6}
	root = buildTree(nums)
	fmt.Println(isValidBST(root))
}

 代码3: 中序遍历+判断

package main

import (
	"fmt"
)

const null = -1 << 31

type TreeNode struct {
	Val   int
	Left  *TreeNode
	Right *TreeNode
}

func isValidBST(root *TreeNode) bool {
	var pre *TreeNode
	stack := []*TreeNode{}
	for root != nil || len(stack) > 0 {
		for root != nil {
			stack = append(stack, root)
			root = root.Left
		}
		node := stack[len(stack)-1]
		stack = stack[:len(stack)-1]
		if pre != nil && node.Val <= pre.Val {
			return false
		}
		pre = node
		root = node.Right
	}
	return true
}

func buildTree(nums []int) *TreeNode {
	if len(nums) == 0 {
		return nil
	}
	root := &TreeNode{Val: nums[0]}
	Queue := []*TreeNode{root}
	idx := 1
	for idx < len(nums) {
		node := Queue[0]
		Queue = Queue[1:]
		if nums[idx] != null {
			node.Left = &TreeNode{Val: nums[idx]}
			Queue = append(Queue, node.Left)
		}
		idx++
		if idx < len(nums) && nums[idx] != null {
			node.Right = &TreeNode{Val: nums[idx]}
			Queue = append(Queue, node.Right)
		}
		idx++
	}
	return root
}

func main() {
	nums := []int{2, 1, 3}
	root := buildTree(nums)
	fmt.Println(isValidBST(root))
	nums = []int{5, 1, 4, null, null, 3, 6}
	root = buildTree(nums)
	fmt.Println(isValidBST(root))
	nums = []int{3, 1, 5, null, null, 4, 6}
	root = buildTree(nums)
	fmt.Println(isValidBST(root))
}

99. 恢复二叉搜索树 Recover Binary Search Tree

给你二叉搜索树的根节点 root ,该树中的 恰好 两个节点的值被错误地交换。请在不改变其结构的情况下,恢复这棵树 

示例 1:

Golang每日一练(leetDay0033)_第4张图片

输入:root = [1,3,null,null,2]
输出:[3,1,null,null,2]
解释:3 不能是 1 的左孩子,因为 3 > 1 。交换 1 和 3 使二叉搜索树有效。

示例 2:

Golang每日一练(leetDay0033)_第5张图片

输入:root = [3,1,4,null,null,2]
输出:[2,1,4,null,null,3]
解释:2 不能在 3 的右子树中,因为 2 < 3 。交换 2 和 3 使二叉搜索树有效。

提示:

  • 树上节点的数目在范围 [2, 1000] 内
  • -2^31 <= Node.val <= 2^31 - 1

进阶:使用 O(n) 空间复杂度的解法很容易实现。你能想出一个只使用 O(1) 空间的解决方案吗?

代码1:中序遍历+交换节点值

对于二叉搜索树,中序遍历得到的序列是递增的。因此,如果有两个节点的值被错误地交换了,那么在中序遍历序列中一定存在两个相邻的逆序对。具体做法是,在中序遍历的过程中,用一个变量pre记录上一个遍历到的节点,每次遍历到一个节点时,判断其值是否小于pre的值,如果小于,则说明存在逆序对,记录下这两个节点,并继续遍历。最后,交换这两个节点的值即可。

package main

import (
	"fmt"
	"strconv"
)

const null = -1 << 31

type TreeNode struct {
	Val   int
	Left  *TreeNode
	Right *TreeNode
}

func recoverTree(root *TreeNode) {
	var pre, first, second *TreeNode
	var stack []*TreeNode
	for root != nil || len(stack) > 0 {
		for root != nil {
			stack = append(stack, root)
			root = root.Left
		}
		node := stack[len(stack)-1]
		stack = stack[:len(stack)-1]
		if pre != nil && node.Val < pre.Val {
			if first == nil {
				first = pre
			}
			second = node
		}
		pre = node
		root = node.Right
	}
	first.Val, second.Val = second.Val, first.Val
}

func buildTree(nums []int) *TreeNode {
	if len(nums) == 0 {
		return nil
	}
	root := &TreeNode{Val: nums[0]}
	Queue := []*TreeNode{root}
	idx := 1
	for idx < len(nums) {
		node := Queue[0]
		Queue = Queue[1:]
		if nums[idx] != null {
			node.Left = &TreeNode{Val: nums[idx]}
			Queue = append(Queue, node.Left)
		}
		idx++
		if idx < len(nums) && nums[idx] != null {
			node.Right = &TreeNode{Val: nums[idx]}
			Queue = append(Queue, node.Right)
		}
		idx++
	}
	return root
}

func levelOrder(root *TreeNode) string {
	if root == nil {
		return "[]"
	}
	arr := []int{}
	que := []*TreeNode{root}
	for len(que) > 0 {
		levelSize := len(que)
		for i := 0; i < levelSize; i++ {
			node := que[0]
			que = que[1:]
			if node == nil {
				arr = append(arr, null)
				continue
			}
			arr = append(arr, node.Val)
			que = append(que, node.Left, node.Right)
		}
	}
	size := len(arr)
	for size > 0 && arr[size-1] == null {
		arr = arr[:size-1]
		size = len(arr)
	}
	result := "["
	for i, n := range arr {
		if n == null {
			result += "null"
		} else {
			result += strconv.FormatInt(int64(n), 10)
		}
		if i < size-1 {
			result += ","
		} else {
			result += "]"
		}
	}
	return result
}

func main() {
	nums := []int{1, 3, null, null, 2}
	root := buildTree(nums)
	fmt.Println(levelOrder(root))
	recoverTree(root)
	fmt.Println(levelOrder(root))

	nums = []int{3, 1, 4, null, null, 2}
	root = buildTree(nums)
	fmt.Println(levelOrder(root))
	recoverTree(root)
	fmt.Println(levelOrder(root))
}

输出:

[1,3,null,null,2]
[3,1,null,null,2]

[3,1,4,null,null,2]
[2,1,4,null,null,3]

代码2:Morris遍历

Morris遍历是一种不需要额外空间的遍历二叉树的方法,它的核心思想是利用叶子节点的空指针来存储遍历中的临时信息。对于二叉搜索树,Morris中序遍历的过程中,每个节点的左子树都已经被遍历完毕,因此可以在遍历到每个节点时,比较它的值和它的前驱节点的值,如果它的值小于前驱节点的值,那么就找到了一个逆序对。我们用两个指针first和second来记录这两个节点,最后交换它们的值即可。

package main

import (
	"fmt"
	"strconv"
)

const null = -1 << 31

type TreeNode struct {
	Val   int
	Left  *TreeNode
	Right *TreeNode
}

func recoverTree(root *TreeNode) {
	var first, second, pre *TreeNode
	var temp *TreeNode
	for root != nil {
		if root.Left != nil {
			temp = root.Left
			for temp.Right != nil && temp.Right != root {
				temp = temp.Right
			}
			if temp.Right == nil {
				temp.Right = root
				root = root.Left
			} else {
				if pre != nil && root.Val < pre.Val {
					if first == nil {
						first = pre
					}
					second = root
				}
				pre = root
				temp.Right = nil
				root = root.Right
			}
		} else {
			if pre != nil && root.Val < pre.Val {
				if first == nil {
					first = pre
				}
				second = root
			}
			pre = root
			root = root.Right
		}
	}
	first.Val, second.Val = second.Val, first.Val
}

func buildTree(nums []int) *TreeNode {
	if len(nums) == 0 {
		return nil
	}
	root := &TreeNode{Val: nums[0]}
	Queue := []*TreeNode{root}
	idx := 1
	for idx < len(nums) {
		node := Queue[0]
		Queue = Queue[1:]
		if nums[idx] != null {
			node.Left = &TreeNode{Val: nums[idx]}
			Queue = append(Queue, node.Left)
		}
		idx++
		if idx < len(nums) && nums[idx] != null {
			node.Right = &TreeNode{Val: nums[idx]}
			Queue = append(Queue, node.Right)
		}
		idx++
	}
	return root
}

func levelOrder(root *TreeNode) string {
	if root == nil {
		return "[]"
	}
	arr := []int{}
	que := []*TreeNode{root}
	for len(que) > 0 {
		levelSize := len(que)
		for i := 0; i < levelSize; i++ {
			node := que[0]
			que = que[1:]
			if node == nil {
				arr = append(arr, null)
				continue
			}
			arr = append(arr, node.Val)
			que = append(que, node.Left, node.Right)
		}
	}
	size := len(arr)
	for size > 0 && arr[size-1] == null {
		arr = arr[:size-1]
		size = len(arr)
	}
	result := "["
	for i, n := range arr {
		if n == null {
			result += "null"
		} else {
			result += strconv.FormatInt(int64(n), 10)
		}
		if i < size-1 {
			result += ","
		} else {
			result += "]"
		}
	}
	return result
}

func main() {
	nums := []int{1, 3, null, null, 2}
	root := buildTree(nums)
	fmt.Println(levelOrder(root))
	recoverTree(root)
	fmt.Println(levelOrder(root))

	nums = []int{3, 1, 4, null, null, 2}
	root = buildTree(nums)
	fmt.Println(levelOrder(root))
	recoverTree(root)
	fmt.Println(levelOrder(root))
}

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