给定两个大小为N 的排序数组A[]和B[] ，任务是检查是否可以将两个给定的排序数组合并为一个新的排序数组，使得没有两个连续的元素来自同一个数组。
例子：
Input: A[] = {3, 5, 8}, B[] = {2, 4, 6}
Output: Yes
Explanation: Merged array = {B[0], A[0], B[1], A[1], B[2], A[2]}
Since the resultant array is sorted array, the required output is Yes.
Input: A[] = {12, 4, 2, 5, 3}, B[] = {32, 13, 43, 10, 8}
Output: No
处理方法：按照以下步骤解决问题：
- 初始化一个变量，比如flag = true来检查是否可以通过合并给定的两个排序数组来形成一个新的排序数组，这样没有两个连续的元素来自同一个数组。
- 初始化一个变量，比如prev来检查合并数组的前一个元素是来自数组A[]还是数组B[] 。如果prev == 1则前一个元素来自数组A[] ，如果prev == 0那么前一个元素来自数组B[] 。
- 使用变量i和j遍历数组并检查以下条件：
- 如果A [I] 则递增i的值，并更新prev的值设置为0。
- 如果B [j]的则递增j的值，并更新prev的值设置为1。
- 如果A [1] == B [j]与一个先前！= 1，则递增j的值，并更新prev的值设置为1。
- 如果A [1] == B [j]与分组！= 0，则递增i的值，并更新prev的值设置为0。
- 如果以上条件都不满足，则 update flag = false 。
- 最后，打印flag的值。
下面是上述方法的实现：
C++
// C++ program to implement
// the above approach
#include
using namespace std;
// Function to check if it is possible to merge
// the two given arrays with given conditions
bool checkIfPossibleMerge(int A[], int B[], int N)
{
// Stores index of
// the array A[]
int i = 0;
// Stores index of
// the array B[]
int j = 0;
// Check if the previous element are from
// the array A[] or from the array B[]
int prev = -1;
// Check if it is possible to merge the two
// given sorted arrays with given conditions
int flag = 1;
// Traverse both the arrays
while (i < N && j < N) {
// If A[i] is less than B[j] and
// previous element are not from A[]
if (A[i] < B[j] && prev != 0) {
// Update prev
prev = 0;
// Update i
i++;
}
// If B[j] is less than A[i] and
// previous element are not from B[]
else if (B[j] < A[i] && prev != 1) {
// Update prev
prev = 1;
// Update j
j++;
}
// If A[i] equal to B[j]
else if (A[i] == B[j]) {
// If previous element
// are not from B[]
if (prev != 1) {
// Update prev
prev = 1;
// Update j
j++;
}
// If previous element is
// not from A[]
else {
// Update prev
prev = 0;
// Update i
i++;
}
}
// If it is not possible to merge two
// sorted arrays with given conditions
else {
// Update flag
flag = 0;
break;
}
}
return flag;
}
// Driver Code
int main()
{
int A[3] = { 3, 5, 8 };
int B[3] = { 2, 4, 6 };
int N = sizeof(A) / sizeof(A[0]);
if (checkIfPossibleMerge(A, B, N)) {
cout << "Yes";
}
else {
cout << "No";
}
return 0;
} Java
// Java program to implement
// the above approach
import java.io.*;
class GFG{
// Function to check if it is possible to merge
// the two given arrays with given conditions
static boolean checkIfPossibleMerge(int[] A, int[] B,
int N)
{
// Stores index of
// the array A[]
int i = 0;
// Stores index of
// the array B[]
int j = 0;
// Check if the previous element are from
// the array A[] or from the array B[]
int prev = -1;
// Check if it is possible to merge the two
// given sorted arrays with given conditions
boolean flag = true;
// Traverse both the arrays
while (i < N && j < N)
{
// If A[i] is less than B[j] and
// previous element are not from A[]
if (A[i] < B[j] && prev != 0)
{
// Update prev
prev = 0;
// Update i
i++;
}
// If B[j] is less than A[i] and
// previous element are not from B[]
else if (B[j] < A[i] && prev != 1)
{
// Update prev
prev = 1;
// Update j
j++;
}
// If A[i] equal to B[j]
else if (A[i] == B[j])
{
// If previous element
// are not from B[]
if (prev != 1)
{
// Update prev
prev = 1;
// Update j
j++;
}
// If previous element is
// not from A[]
else
{
// Update prev
prev = 0;
// Update i
i++;
}
}
// If it is not possible to merge two
// sorted arrays with given conditions
else
{
// Update flag
flag = false;
break;
}
}
return flag;
}
// Driver Code
public static void main(String[] args)
{
int[] A = { 3, 5, 8 };
int[] B = { 2, 4, 6 };
int N = A.length;
if (checkIfPossibleMerge(A, B, N))
{
System.out.println("Yes");
}
else
{
System.out.println("No");
}
}
}
// This code is contributed by akhilsainiPython3
# Python3 program to implement
# the above approach
# Function to check if it is possible
# to merge the two given arrays with
# given conditions
def checkIfPossibleMerge(A, B, N):
# Stores index of
# the array A[]
i = 0
# Stores index of
# the array B[]
j = 0
# Check if the previous element
# are from the array A[] or from
# the array B[]
prev = -1
# Check if it is possible to merge
# the two given sorted arrays with
# given conditions
flag = 1
# Traverse both the arrays
while (i < N and j < N):
# If A[i] is less than B[j] and
# previous element are not from A[]
if (A[i] < B[j] and prev != 0):
# Update prev
prev = 0
# Update i
i += 1
# If B[j] is less than A[i] and
# previous element are not from B[]
elif (B[j] < A[i] and prev != 1):
# Update prev
prev = 1
# Update j
j += 1
# If A[i] equal to B[j]
elif (A[i] == B[j]):
# If previous element
# are not from B[]
if (prev != 1):
# Update prev
prev = 1
# Update j
j += 1
# If previous element is
# not from A[]
else:
# Update prev
prev = 0
# Update i
i += 1
# If it is not possible to merge two
# sorted arrays with given conditions
else:
# Update flag
flag = 0
break
return flag
# Driver Code
if __name__ == '__main__':
A = [ 3, 5, 8 ]
B = [ 2, 4, 6 ]
N = len(A)
if (checkIfPossibleMerge(A, B, N)):
print("Yes")
else:
print("No")
# This code is contributed by akhilsainiC#
// C# program to implement
// the above approach
using System;
class GFG{
// Function to check if it is possible to merge
// the two given arrays with given conditions
static bool checkIfPossibleMerge(int[] A, int[] B,
int N)
{
// Stores index of
// the array A[]
int i = 0;
// Stores index of
// the array B[]
int j = 0;
// Check if the previous element are
// from the array A[] or from the
// array B[]
int prev = -1;
// Check if it is possible to merge
// the two given sorted arrays with
// given conditions
bool flag = true;
// Traverse both the arrays
while (i < N && j < N)
{
// If A[i] is less than B[j] and
// previous element are not from A[]
if (A[i] < B[j] && prev != 0)
{
// Update prev
prev = 0;
// Update i
i++;
}
// If B[j] is less than A[i] and
// previous element are not from B[]
else if (B[j] < A[i] && prev != 1)
{
// Update prev
prev = 1;
// Update j
j++;
}
// If A[i] equal to B[j]
else if (A[i] == B[j])
{
// If previous element
// are not from B[]
if (prev != 1)
{
// Update prev
prev = 1;
// Update j
j++;
}
// If previous element is
// not from A[]
else
{
// Update prev
prev = 0;
// Update i
i++;
}
}
// If it is not possible to merge two
// sorted arrays with given conditions
else
{
// Update flag
flag = false;
break;
}
}
return flag;
}
// Driver Code
public static void Main()
{
int[] A = { 3, 5, 8 };
int[] B = { 2, 4, 6 };
int N = A.Length;
if (checkIfPossibleMerge(A, B, N))
{
Console.WriteLine("Yes");
}
else
{
Console.WriteLine("No");
}
}
}
// This code is contributed by akhilsainiJavascript
输出：
Yes时间复杂度： O(N)
辅助空间： O(1)
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