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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给定三个字符串 s1
、s2
、s3
,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
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:
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
s1
、s2
、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))
}
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:
示例 1:
输入:root = [2,1,3] 输出:true
示例 2:
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:
[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))
}
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:
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:
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:
[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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Golang每日一练 专栏 |
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