原文：
给定一个由n个整数和一个整数k组成的数组，请选择两个总和为k的不同元素，并找出所选择元素到端点的最大最短距离。
示例：
input : a[] = {2, 4, 3, 2, 1}
k = 5.
output : 2
explanation:
select the pair(4, 1).
shortest distance of 4 from ends = 2
shortest distance of 1 from ends = 1
hence, answer is max(2, 1) = 2
input : a[] = {2, 4, 1, 9, 5}
k = 3
output : 3
explanation:
select the pair (2, 1)
shortest distance of 2 from ends = 1
shortest distance of 1 from ends = 3
hence, answer is max(1, 3) = 3\.
注意：末端元素到末端的距离是 1 而不是 0。
朴素的方法：该方法是运行两个循环，并在内部循环中检查两个元素是否与和k成对。 如果是，则以两个元素之间最短距离的最大值作为答案，将其与前一对元素的最短距离进行比较，并以这两个元素中的最小值作为答案。 循环结束时，我们将获得所需的输出。
有效方法：显然，最短距离是指距左端的距离和距右端的距离，即min(1 i, n-i)。 让我们将第 i 个元素的最短距离表示为di。 还有另一种情况，即重复选定对中的一个元素，然后在该元素出现的所有最短距离中选择最小的一个。 循环运行，并将所有数组元素中最短的距离存储在另一个数组中（将其设为d[]）。 现在，我们得到了所有元素的最短距离。
运行for循环。 如果选取的元素是x，则另一个元素应该是k-x。 用max(d[x], d[k-x])更新ans，并在每次更新时，选择先前和当前答案中的最小值。 如果k-x不在数组中，则d[k-x]将为infinite，这将已经初始化。
c
// c code to find maximum shortest distance
// from endpoints
#include
using namespace std;
// function to find maximum shortest distance
int find_maximum(int a[], int n, int k)
{
// stores the shortest distance of every
// element in original array.
unordered_map b;
for (int i = 0; i < n; i ) {
int x = a[i];
// shortest distance from ends
int d = min(1 i, n - i);
if (b.find(x) == b.end())
b[x] = d;
else
/* if duplicates are found, b[x]
is replaced with minimum of the
previous and current position's
shortest distance*/
b[x] = min(d, b[x]);
}
int ans = int_max;
for (int i = 0; i < n; i ) {
int x = a[i];
// similar elements ignore them
// cause we need distinct elements
if (x != k - x && b.find(k - x) != b.end())
ans = min(max(b[x], b[k - x]), ans);
}
return ans;
}
// driver code
int main()
{
int a[] = { 3, 5, 8, 6, 7 };
int k = 11;
int n = sizeof(a) / sizeof(a[0]);
cout << find_maximum(a, n, k) << endl;
return 0;
}
java
// java code to find maximum shortest distance
// from endpoints
import java.util.*;
class gfg
{
static void makepermutation(int []a, int n)
{
// store counts of all elements.
hashmap count = new hashmap();
for (int i = 0; i < n; i )
{
if(count.containskey(a[i]))
{
count.put(a[i], count.get(a[i]) 1);
}
else
{
count.put(a[i], 1);
}
}
}
// function to find maximum shortest distance
static int find_maximum(int a[], int n, int k)
{
// stores the shortest distance of every
// element in original array.
hashmap b = new hashmap();
for (int i = 0; i < n; i )
{
int x = a[i];
// shortest distance from ends
int d = math.min(1 i, n - i);
if (!b.containskey(x))
b.put(x, d);
else
{
/* if duplicates are found, b[x]
is replaced with minimum of the
previous and current position's
shortest distance*/
b. put(x, math.min(d, b.get(x)));
}
}
int ans = integer.max_value;
for (int i = 0; i < n; i )
{
int x = a[i];
// similar elements ignore them
// cause we need distinct elements
if (x != k - x && b.containskey(k - x))
ans = math.min(math.max(b.get(x),
b.get(k - x)), ans);
}
return ans;
}
// driver code
public static void main(string[] args)
{
int a[] = { 3, 5, 8, 6, 7 };
int k = 11;
int n = a.length;
system.out.println(find_maximum(a, n, k));
}
}
// this code is contributed by rajput-ji
python3
# python3 code to find maximum shortest
# distance from endpoints
# function to find maximum shortest distance
def find_maximum(a, n, k):
# stores the shortest distance of every
# element in original array.
b = dict()
for i in range(n):
x = a[i]
# shortest distance from ends
d = min(1 i, n - i)
if x not in b.keys():
b[x] = d
else:
# if duplicates are found, b[x]
# is replaced with minimum of the
# previous and current position's
# shortest distance*/
b[x] = min(d, b[x])
ans = 10**9
for i in range(n):
x = a[i]
# similar elements ignore them
# cause we need distinct elements
if (x != (k - x) and (k - x) in b.keys()):
ans = min(max(b[x], b[k - x]), ans)
return ans
# driver code
a = [3, 5, 8, 6, 7]
k = 11
n = len(a)
print(find_maximum(a, n, k))
# this code is contributed by mohit kumar
c
// c# code to find maximum shortest distance
// from endpoints
using system;
using system.collections.generic;
class gfg
{
static void makepermutation(int []a, int n)
{
// store counts of all elements.
dictionary count = new dictionary();
for (int i = 0; i < n; i )
{
if(count.containskey(a[i]))
{
count[a[i]] = count[a[i]] 1;
}
else
{
count.add(a[i], 1);
}
}
}
// function to find maximum shortest distance
static int find_maximum(int []a, int n, int k)
{
// stores the shortest distance of every
// element in original array.
dictionary b = new dictionary();
for (int i = 0; i < n; i )
{
int x = a[i];
// shortest distance from ends
int d = math.min(1 i, n - i);
if (!b.containskey(x))
b.add(x, d);
else
{
/* if duplicates are found, b[x]
is replaced with minimum of the
previous and current position's
shortest distance*/
b[x] = math.min(d, b[x]);
}
}
int ans = int.maxvalue;
for (int i = 0; i < n; i )
{
int x = a[i];
// similar elements ignore them
// cause we need distinct elements
if (x != k - x && b.containskey(k - x))
ans = math.min(math.max(b[x],
b[k - x]), ans);
}
return ans;
}
// driver code
public static void main(string[] args)
{
int []a = { 3, 5, 8, 6, 7 };
int k = 11;
int n = a.length;
console.writeline(find_maximum(a, n, k));
}
}
// this code is contributed by princi singh
输出：
2
麻将胡了pg电子网站的版权属于：月萌api www.moonapi.com，转载请注明出处
