给定一个由N个整数组成的数组A [] ，任务是查找给定数组中每个元素的相对等级。
The relative rank for each element in the array is the count of elements which is greater than the current element in the Longest Increasing Subsequence from the current element.
例子：
Input: A[] = {8, 16, 5, 6, 9}, N = 5
Output: {1, 0, 2, 1, 0}
Explanation:
For i = 0, required sequence is {8, 16} Relative Rank = 1.
For i = 1, Since all elements after 16 are smaller than 16, Relative Rank = 0.
For i = 2, required sequence is {5, 6, 9} Relative Rank = 2
For i = 3, required sequence is {6, 9} Relative Rank = 1
For i = 4, required sequence is {9} Relative Rank = 0
Input: A[] = {1, 2, 3, 5, 4}
Output: {3, 2, 1, 0, 0}
Explanation:
For i = 0, required sequence is {1, 2, 3, 5}, Relative Rank = 3
For i = 1, required sequence is {2, 3, 5}, Relative Rank = 2
For i = 2, required sequence is {3, 5}, Relative Rank = 1
For i = 3, required sequence is {5}, Relative Rank = 0
For i = 4, required sequence is {4}, Relative Rank = 0
天真的方法：这个想法是为每个元素生成最长的递增子序列，然后，每个元素的相对等级为(LIS – 1的长度) 。
时间复杂度： O(N 2 )
辅助空间： O(1)
高效方法：为了优化上述方法，我们的想法是使用堆栈并存储从右到每个元素(例如A [i] )以非降序排列的元素，然后每个A [i]的排名为(堆栈大小– 1)直到该元素。下面是相同的插图： 
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to find relative rank for
// each element in the array A[]
void findRank(int A[], int N)
{
// Create Rank Array
int rank[N] = {};
// Stack to store numbers in
// non-decreasing order from right
stack s;
// Push last element in stack
s.push(A[N - 1]);
// Iterate from second last
// element to first element
for (int i = N - 2; i >= 0; i--) {
// If current element is less
// than the top of stack and
// push A[i] in stack
if (A[i] < s.top()) {
s.push(A[i]);
// Rank is stack size - 1
// for current element
rank[i] = s.size() - 1;
}
else {
// Pop elements from stack
// till current element is
// greater than the top
while (!s.empty()
&& A[i] >= s.top()) {
s.pop();
}
// Push current element in Stack
s.push(A[i]);
// Rank is stack size - 1
rank[i] = s.size() - 1;
}
}
// Print rank of all elements
for (int i = 0; i < N; i++) {
cout << rank[i] << " ";
}
}
// Driver Code
int main()
{
// Given array A[]
int A[] = { 1, 2, 3, 5, 4 };
int N = sizeof(A) / sizeof(A[0]);
// Function call
findRank(A, N);
return 0;
} Java
// Java program to implement
// the above approach
import java.io.*;
import java.util.*;
import java.lang.*;
class GFG{
// Function to find relative rank for
// each element in the array A[]
static void findRank(int[] A, int N)
{
// Create Rank Array
int[] rank = new int[N];
// Stack to store numbers in
// non-decreasing order from right
Stack s = new Stack();
// Push last element in stack
s.add(A[N - 1]);
// Iterate from second last
// element to first element
for(int i = N - 2; i >= 0; i--)
{
// If current element is less
// than the top of stack and
// push A[i] in stack
if (A[i] < s.peek())
{
s.add(A[i]);
// Rank is stack size - 1
// for current element
rank[i] = s.size() - 1;
}
else
{
// Pop elements from stack
// till current element is
// greater than the top
while (!s.isEmpty() &&
A[i] >= s.peek())
{
s.pop();
}
// Push current element in Stack
s.add(A[i]);
// Rank is stack size - 1
rank[i] = s.size() - 1;
}
}
// Print rank of all elements
for(int i = 0; i < N; i++)
{
System.out.print(rank[i] + " ");
}
}
// Driver Code
public static void main(String[] args)
{
// Given array A[]
int A[] = { 1, 2, 3, 5, 4 };
int N = A.length;
// Function call
findRank(A, N);
}
}
// This code is contributed by sanjoy_62 Python3
# Python3 program for the above approach
# Function to find relative rank for
# each element in the array A[]
def findRank(A, N):
# Create Rank Array
rank = [0] * N
# Stack to store numbers in
# non-decreasing order from right
s = []
# Push last element in stack
s.append(A[N - 1])
# Iterate from second last
# element to first element
for i in range(N - 2, -1, -1):
# If current element is less
# than the top of stack and
# append A[i] in stack
if (A[i] < s[-1]):
s.append(A[i])
# Rank is stack size - 1
# for current element
rank[i] = len(s) - 1
else:
# Pop elements from stack
# till current element is
# greater than the top
while (len(s) > 0 and A[i] >= s[-1]):
del s[-1]
# Push current element in Stack
s.append(A[i])
# Rank is stack size - 1
rank[i] = len(s) - 1
# Print rank of all elements
for i in range(N):
print(rank[i], end = " ")
# Driver Code
if __name__ == '__main__':
# Given array A[]
A = [ 1, 2, 3, 5, 4 ]
N = len(A)
# Function call
findRank(A, N)
# This code is contributed by mohit kumar 29C#
// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
using System.Linq;
class GFG{
// Function to find relative rank for
// each element in the array A[]
static void findRank(int[] A, int N)
{
// Create Rank Array
int[] rank = new int[N];
// Stack to store numbers in
// non-decreasing order from right
Stack s = new Stack();
// Push last element in stack
s.Push(A[N - 1]);
// Iterate from second last
// element to first element
for(int i = N - 2; i >= 0; i--)
{
// If current element is less
// than the top of stack and
// push A[i] in stack
if (A[i] < s.Peek())
{
s.Push(A[i]);
// Rank is stack size - 1
// for current element
rank[i] = s.Count() - 1;
}
else
{
// Pop elements from stack
// till current element is
// greater than the top
while (s.Count() != 0 &&
A[i] >= s.Peek())
{
s.Pop();
}
// Push current element in Stack
s.Push(A[i]);
// Rank is stack size - 1
rank[i] = s.Count() - 1;
}
}
// Print rank of all elements
for(int i = 0; i < N; i++)
{
Console.Write(rank[i] + " ");
}
}
// Driver Code
public static void Main()
{
// Given array A[]
int[] A = new int[] { 1, 2, 3, 5, 4 };
int N = A.Length;
// Function call
findRank(A, N);
}
}
// This code is contributed by sanjoy_62 3 2 1 0 0
时间复杂度： O(N)
辅助空间： O(1)
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