在 2D 阵列中找到一个峰值元素
原文:https://www.geeksforgeeks.org/find-peak-element-2d-array/
如果一个元素大于或等于它的四个相邻元素,即左、右、上和下,则它是一个峰值元素。例如,A[i][j]的邻居是 A[i-1][j]、A[i+1][j]、A[i][j-1]和 A[i][j+1]。对于角元素,缺少的邻居被认为是负无穷大值。 例:
Input : 10 20 15
21 30 14
7 16 32
Output : 30
30 is a peak element because all its
neighbors are smaller or equal to it.
32 can also be picked as a peak.
Input : 10 7
11 17
Output : 17
以下是关于这个问题的一些事实: 1:对角线相邻不被认为是邻居。 2:一个峰元素不一定是最大元素。 3:可以存在不止一种这样的元素。 4:总有峰元素。我们可以通过使用笔和纸创建一些矩阵来看到这个属性。 方法一:(蛮力) 迭代矩阵的所有元素,检查它是否大于/等于它的所有邻居。如果是,返回元素。 时间复杂度:O(行列) 辅助空间:O(1) 方法二:(高效)* 这个问题主要是在 1D 阵中找峰元素的一个延伸。我们在此应用类似的基于二分搜索法的解决方案。
- 考虑中柱,并在其中找到最大元素。
-
让中间列的索引为“中间”,中间列中最大元素的值为“最大”,最大元素位于“mat[max_index][mid]”。
-
如果最大值> = A[索引][中间 1] &最大值> = A[索引][选择+1],则最大值为峰值,返回最大值。
- 如果 max < mat[max_index][mid-1],则对矩阵的左半部分重复。
- 如果 max < mat[max_index][mid+1],则对矩阵的右半部分重复。
以下是上述算法的实现:
C++
// Finding peak element in a 2D Array.
#include <bits/stdc++.h>
using namespace std;
const int MAX = 100;
// Function to find the maximum in column 'mid'
// 'rows' is number of rows.
int findMax(int arr[][MAX], int rows, int mid, int& max)
{
int max_index = 0;
for (int i = 0; i < rows; i++) {
if (max < arr[i][mid]) {
// Saving global maximum and its index
// to check its neighbours
max = arr[i][mid];
max_index = i;
}
}
return max_index;
}
// Function to find a peak element
int findPeakRec(int arr[][MAX], int rows, int columns,
int mid)
{
// Evaluating maximum of mid column. Note max is
// passed by reference.
int max = 0;
int max_index = findMax(arr, rows, mid, max);
// If we are on the first or last column,
// max is a peak
if (mid == 0 || mid == columns - 1)
return max;
// If mid column maximum is also peak
if (max >= arr[max_index][mid - 1] && max >= arr[max_index][mid + 1])
return max;
// If max is less than its left
if (max < arr[max_index][mid - 1])
return findPeakRec(arr, rows, columns, mid - ceil((double)mid / 2));
// If max is less than its left
// if (max < arr[max_index][mid+1])
return findPeakRec(arr, rows, columns, mid + ceil((double)mid / 2));
}
// A wrapper over findPeakRec()
int findPeak(int arr[][MAX], int rows, int columns)
{
return findPeakRec(arr, rows, columns, columns / 2);
}
// Driver Code
int main()
{
int arr[][MAX] = { { 10, 8, 10, 10 },
{ 14, 13, 12, 11 },
{ 15, 9, 11, 21 },
{ 16, 17, 19, 20 } };
// Number of Columns
int rows = 4, columns = 4;
cout << findPeak(arr, rows, columns);
return 0;
}
Java 语言(一种计算机语言,尤用于创建网站)
// Finding peak element in a 2D Array.
class GFG
{
static int MAX = 100;
// Function to find the maximum in column
// 'mid', 'rows' is number of rows.
static int findMax(int[][] arr, int rows,
int mid, int max)
{
int max_index = 0;
for (int i = 0; i < rows; i++)
{
if (max < arr[i][mid])
{
// Saving global maximum and its index
// to check its neighbours
max = arr[i][mid];
max_index = i;
}
}
return max_index;
}
// Function to change the value of [max]
static int Max(int[][] arr, int rows,
int mid, int max)
{
for (int i = 0; i < rows; i++)
{
if (max < arr[i][mid])
{
// Saving global maximum and its index
// to check its neighbours
max = arr[i][mid];
}
}
return max;
}
// Function to find a peak element
static int findPeakRec(int[][] arr, int rows,
int columns, int mid)
{
// Evaluating maximum of mid column.
// Note max is passed by reference.
int max = 0;
int max_index = findMax(arr, rows, mid, max);
max = Max(arr, rows, mid, max);
// If we are on the first or last column,
// max is a peak
if (mid == 0 || mid == columns - 1)
return max;
// If mid column maximum is also peak
if (max >= arr[max_index][mid - 1] &&
max >= arr[max_index][mid + 1])
return max;
// If max is less than its left
if (max < arr[max_index][mid - 1])
return findPeakRec(arr, rows, columns,
(int)(mid - Math.ceil((double) mid / 2)));
// If max is less than its left
// if (max < arr[max_index][mid+1])
return findPeakRec(arr, rows, columns,
(int)(mid + Math.ceil((double) mid / 2)));
}
// A wrapper over findPeakRec()
static int findPeak(int[][] arr, int rows, int columns)
{
return findPeakRec(arr, rows, columns, columns / 2);
}
// Driver Code
public static void main(String[] args)
{
int[][] arr = {{ 10, 8, 10, 10 },
{ 14, 13, 12, 11 },
{ 15, 9, 11, 21 },
{ 16, 17, 19, 20 }};
// Number of Columns
int rows = 4, columns = 4;
System.out.println(findPeak(arr, rows, columns));
}
}
// This code is contributed by
// sanjeev2552
Python 3
# Finding peak element in a 2D Array.
MAX = 100
from math import ceil
# Function to find the maximum in column 'mid'
# 'rows' is number of rows.
def findMax(arr, rows, mid,max):
max_index = 0
for i in range(rows):
if (max < arr[i][mid]):
# Saving global maximum and its index
# to check its neighbours
max = arr[i][mid]
max_index = i
#print(max_index)
return max,max_index
# Function to find a peak element
def findPeakRec(arr, rows, columns,mid):
# Evaluating maximum of mid column.
# Note max is passed by reference.
max = 0
max, max_index = findMax(arr, rows, mid, max)
# If we are on the first or last column,
# max is a peak
if (mid == 0 or mid == columns - 1):
return max
# If mid column maximum is also peak
if (max >= arr[max_index][mid - 1] and
max >= arr[max_index][mid + 1]):
return max
# If max is less than its left
if (max < arr[max_index][mid - 1]):
return findPeakRec(arr, rows, columns,
mid - ceil(mid / 2.0))
# If max is less than its left
# if (max < arr[max_index][mid+1])
return findPeakRec(arr, rows, columns,
mid + ceil(mid / 2.0))
# A wrapper over findPeakRec()
def findPeak(arr, rows, columns):
return findPeakRec(arr, rows,
columns, columns // 2)
# Driver Code
arr = [ [ 10, 8, 10, 10 ],
[ 14, 13, 12, 11 ],
[ 15, 9, 11, 21 ],
[ 16, 17, 19, 20 ] ]
# Number of Columns
rows = 4
columns = 4
print(findPeak(arr, rows, columns))
# This code is contributed by Mohit Kumar
C
// Finding peak element in a 2D Array.
using System;
class GFG
{
// Function to find the maximum in column
// 'mid', 'rows' is number of rows.
static int findMax(int[,] arr, int rows,
int mid, int max)
{
int max_index = 0;
for (int i = 0; i < rows; i++)
{
if (max < arr[i,mid])
{
// Saving global maximum and its
// index to check its neighbours
max = arr[i,mid];
max_index = i;
}
}
return max_index;
}
// Function to change the value of [max]
static int Max(int[,] arr, int rows,
int mid, int max)
{
for (int i = 0; i < rows; i++)
{
if (max < arr[i, mid])
{
// Saving global maximum and its
// index to check its neighbours
max = arr[i, mid];
}
}
return max;
}
// Function to find a peak element
static int findPeakRec(int[,] arr, int rows,
int columns, int mid)
{
// Evaluating maximum of mid column.
// Note max is passed by reference.
int max = 0;
int max_index = findMax(arr, rows, mid, max);
max = Max(arr, rows, mid, max);
// If we are on the first or last column,
// max is a peak
if (mid == 0 || mid == columns - 1)
return max;
// If mid column maximum is also peak
if (max >= arr[max_index, mid - 1] &&
max >= arr[max_index, mid + 1])
return max;
// If max is less than its left
if (max < arr[max_index,mid - 1])
return findPeakRec(arr, rows, columns,
(int)(mid - Math.Ceiling((double) mid / 2)));
// If max is less than its left
// if (max < arr[max_index][mid+1])
return findPeakRec(arr, rows, columns,
(int)(mid + Math.Ceiling((double) mid / 2)));
}
// A wrapper over findPeakRec()
static int findPeak(int[,] arr,
int rows, int columns)
{
return findPeakRec(arr, rows, columns,
columns / 2);
}
// Driver Code
static public void Main ()
{
int[,] arr = {{ 10, 8, 10, 10 },
{ 14, 13, 12, 11 },
{ 15, 9, 11, 21 },
{ 16, 17, 19, 20 }};
// Number of Columns
int rows = 4, columns = 4;
Console.Write(findPeak(arr, rows, columns));
}
}
// This code is contributed by ajit.
java 描述语言
<script>
// Finding peak element in a 2D Array.
let MAX = 100;
// Function to find the maximum in column
// 'mid', 'rows' is number of rows.
function findMax(arr, rows, mid, max)
{
let max_index = 0;
for (let i = 0; i < rows; i++)
{
if (max < arr[i][mid])
{
// Saving global maximum and its index
// to check its neighbours
max = arr[i][mid];
max_index = i;
}
}
return max_index;
}
// Function to change the value of [max]
function Max(arr, rows, mid, max)
{
for (let i = 0; i < rows; i++)
{
if (max < arr[i][mid])
{
// Saving global maximum and its index
// to check its neighbours
max = arr[i][mid];
}
}
return max;
}
// Function to find a peak element
function findPeakRec(arr, rows, columns, mid)
{
// Evaluating maximum of mid column.
// Note max is passed by reference.
let max = 0;
let max_index = findMax(arr, rows, mid, max);
max = Max(arr, rows, mid, max);
// If we are on the first or last column,
// max is a peak
if (mid == 0 || mid == columns - 1)
return max;
// If mid column maximum is also peak
if (max >= arr[max_index][mid - 1] &&
max >= arr[max_index][mid + 1])
return max;
// If max is less than its left
if (max < arr[max_index][mid - 1])
return findPeakRec(arr, rows, columns, (mid - Math.ceil(mid / 2)));
// If max is less than its left
// if (max < arr[max_index][mid+1])
return findPeakRec(arr, rows, columns, (mid + Math.ceil(mid / 2)));
}
// A wrapper over findPeakRec()
function findPeak(arr, rows, columns)
{
return findPeakRec(arr, rows, columns, parseInt(columns / 2, 10));
}
let arr = [[ 10, 8, 10, 10 ],
[ 14, 13, 12, 11 ],
[ 15, 9, 11, 21 ],
[ 16, 17, 19, 20 ]];
// Number of Columns
let rows = 4, columns = 4;
document.write(findPeak(arr, rows, columns));
</script>
输出:
21
时间复杂度:O(行日志(列))。我们重复了一半的列数。在每次递归调用中,我们线性搜索当前中间列中的最大值。 辅助空间:O(列/2)为递归调用栈 来源:* https://OCW . MIT . edu/courses/electric-engineering-and-computer-science/6-006-算法导论-fall-2011/讲座-视频/MIT6_006F11_lec01.pdf
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