给定一个由不同元素组成的数组arr [] ，任务是找到每个可能子数组的第一个和第二个最大元素的XOR值的最大值。
注意：数组的长度大于1。
例子：
Input: arr[] = {5, 4, 3}
Output: 7
Explanation:
All Possible subarrays with length greater than 1 and their XOR values of first and second maximum element –
XOR of First and Second maximum({5, 4}) = 1
XOR of First and Second maximum({5, 4, 3}) = 1
XOR of First and Second maximum({4, 3}) = 7
Input: arr[] = {9, 8, 3, 5, 7}
Output: 15
天真的方法：生成所有可能的长度大于1的子数组，并为每个可能的子数组找到子数组的第一个和第二个最大元素的XOR值，并从中找出最大值。
高效的方法：对于此问题，请保持堆栈并遵循给定的步骤
- 从左到右遍历给定的数组，然后对于每个元素arr [i] –
- 如果堆栈顶部小于arr [i]，则从堆栈中弹出元素，直到堆栈顶部小于arr [i]。
- 将arr [i]推入堆栈。
- 查找堆栈顶部两个元素的XOR值，如果当前XOR值大于找到的最大值，则更新最大值。
下面是上述方法的实现：
C++
// C++ implementation of the above approach.
#include
using namespace std;
// Function to find the maximum XOR value
int findMaxXOR(vector arr, int n){
vector stack;
int res = 0, l = 0, i;
// Traversing given array
for (i = 0; i < n; i++) {
// If there are elements in stack
// and top of stack is less than
// current element then pop the elements
while (!stack.empty() &&
stack.back() < arr[i]) {
stack.pop_back();
l--;
}
// Push current element
stack.push_back(arr[i]);
// Increasing length of stack
l++;
if (l > 1) {
// Updating the maximum result
res = max(res,
stack[l - 1] ^ stack[l - 2]);
}
}
return res;
}
// Driver Code
int main()
{
// Initializing array
vector arr{ 9, 8, 3, 5, 7 };
int result1 = findMaxXOR(arr, 5);
// Reversing the array(vector)
reverse(arr.begin(), arr.end());
int result2 = findMaxXOR(arr, 5);
cout << max(result1, result2);
return 0;
} Java
// Java implementation of the above approach.
import java.util.*;
class GFG{
// Function to find the maximum XOR value
static int findMaxXOR(Vector arr, int n){
Vector stack = new Vector();
int res = 0, l = 0, i;
// Traversing given array
for (i = 0; i < n; i++) {
// If there are elements in stack
// and top of stack is less than
// current element then pop the elements
while (!stack.isEmpty() &&
stack.get(stack.size()-1) < arr.get(i)) {
stack.remove(stack.size()-1);
l--;
}
// Push current element
stack.add(arr.get(i));
// Increasing length of stack
l++;
if (l > 1) {
// Updating the maximum result
res = Math.max(res,
stack.get(l - 1) ^ stack.get(l - 2));
}
}
return res;
}
// Driver Code
public static void main(String[] args)
{
// Initializing array
Integer []temp = { 9, 8, 3, 5, 7 };
Vector arr = new Vector<>(Arrays.asList(temp));
int result1 = findMaxXOR(arr, 5);
// Reversing the array(vector)
Collections.reverse(arr);
int result2 = findMaxXOR(arr, 5);
System.out.print(Math.max(result1, result2));
}
}
// This code is contributed by sapnasingh4991 Python 3
# Python implementation of the approach
from collections import deque
def maxXOR(arr):
# Declaring stack
stack = deque()
# Initializing the length of stack
l = 0
# Initializing res1 for array
# traversal of left to right
res1 = 0
# Traversing the array
for i in arr:
# If there are elements in stack
# And top of stack is less than
# current element then pop the stack
while stack and stack[-1]1:
res1 = max(res1, stack[-1]^stack[-2])
# Similar to the above method,
# we calculate the xor for reversed array
res2 = 0
# Clear the whole stack
stack.clear()
l = 0
# Reversing the array
arr.reverse()
for i in arr:
while stack and stack[-1]1:
res2 = max(res2, stack[-1]^stack[-2])
# Printing the maximum of res1, res2
return max(res1, res2)
# Driver Code
if __name__ == "__main__":
# Initializing the array
arr = [9, 8, 3, 5, 7]
print(maxXOR(arr)) C#
// C# implementation of the above approach.
using System;
using System.Collections.Generic;
class GFG{
// Function to find the maximum XOR value
static int findMaxXOR(List arr, int n){
List stack = new List();
int res = 0, l = 0, i;
// Traversing given array
for (i = 0; i < n; i++) {
// If there are elements in stack
// and top of stack is less than
// current element then pop the elements
while (stack.Count!=0 &&
stack[stack.Count-1] < arr[i]) {
stack.RemoveAt(stack.Count-1);
l--;
}
// Push current element
stack.Add(arr[i]);
// Increasing length of stack
l++;
if (l > 1) {
// Updating the maximum result
res = Math.Max(res,
stack[l - 1] ^ stack[l - 2]);
}
}
return res;
}
// Driver Code
public static void Main(String[] args)
{
// Initializing array
int []temp = { 9, 8, 3, 5, 7 };
List arr = new List(temp);
int result1 = findMaxXOR(arr, 5);
// Reversing the array(vector)
arr.Reverse();
int result2 = findMaxXOR(arr, 5);
Console.Write(Math.Max(result1, result2));
}
}
// This code is contributed by 29AjayKumar 输出：
15