给定一个数组arr []，找到在最多执行k次连续交换后可以得到的字典上最小的数组。
例子 ：
Input: arr[] = {7, 6, 9, 2, 1}
k = 3
Output: arr[] = {2, 7, 6, 9, 1}
Explanation: Array is: 7, 6, 9, 2, 1
Swap 1: 7, 6, 2, 9, 1
Swap 2: 7, 2, 6, 9, 1
Swap 3: 2, 7, 6, 9, 1
So Our final array after k = 3 swaps :
2, 7, 6, 9, 1
Input: arr[] = {7, 6, 9, 2, 1}
k = 1
Output: arr[] = {6, 7, 9, 2, 1}
我们强烈建议您单击此处并进行实践，然后再继续解决方案。
天真的方法是生成数组的所有排列，并选择满足最多k个交换条件的最小排列。此方法的时间复杂度为＆ohm;(n！)，对于较大的n值，肯定会超时。
一种有效的方法是贪婪地思考。我们首先从数组a 1 ，a 2 ，a 3 …(a k或n )中选取最小的元素[当k较小时考虑k ，否则n]。将所有这些元素右移1个位置后，将最小的元素放置在0位置。我们从k中减去掉期数(掉期数是班次数减去1)。如果仍然使k> 0，则从下一个起始位置应用相同的过程，即a 2 ，a 3 …(a k或n )，然后将其放置在a 1位置。因此，我们将继续应用相同的过程，直到k变为0。
C++
// C++ program to find lexicographically minimum
// value after k swaps.
#include
using namespace std ;
// Modifies arr[0..n-1] to lexicographically smallest
// with k swaps.
void minimizeWithKSwaps(int arr[], int n, int k)
{
for (int i = 0; i0; ++i)
{
// Set the position where we want
// to put the smallest integer
int pos = i;
for (int j = i+1; j k)
break;
// Find the minimum value from i+1 to
// max k or n
if (arr[j] < arr[pos])
pos = j;
}
// Swap the elements from Minimum position
// we found till now to the i index
for (int j = pos; j>i; --j)
swap(arr[j], arr[j-1]);
// Set the final value after swapping pos-i
// elements
k -= pos-i;
}
}
// Driver code
int main()
{
int arr[] = {7, 6, 9, 2, 1};
int n = sizeof(arr)/sizeof(arr[0]);
int k = 3;
minimizeWithKSwaps(arr, n, k);
//Print the final Array
for (int i=0; i Java
// Java program to find lexicographically minimum
// value after k swaps.
class GFG {
// Modifies arr[0..n-1] to lexicographically
// smallest with k swaps.
static void minimizeWithKSwaps(int arr[], int n, int k)
{
for (int i = 0; i < n-1 && k > 0; ++i)
{
// Set the position where we want
// to put the smallest integer
int pos = i;
for (int j = i+1; j < n ; ++j)
{
// If we exceed the Max swaps
// then terminate the loop
if (j - i > k)
break;
// Find the minimum value from i+1 to
// max k or n
if (arr[j] < arr[pos])
pos = j;
}
// Swap the elements from Minimum position
// we found till now to the i index
int temp;
for (int j = pos; j>i; --j)
{
temp=arr[j];
arr[j]=arr[j-1];
arr[j-1]=temp;
}
// Set the final value after swapping pos-i
// elements
k -= pos-i;
}
}
// Driver method
public static void main(String[] args)
{
int arr[] = {7, 6, 9, 2, 1};
int n = arr.length;
int k = 3;
minimizeWithKSwaps(arr, n, k);
//Print the final Array
for (int i=0; iPython
# Python program to find lexicographically minimum
# value after k swaps.
def minimizeWithKSwaps(arr, n, k):
for i in range(n-1):
# Set the position where we we want
# to put the smallest integer
pos = i
for j in range(i+1, n):
# If we exceed the Max swaps
# then terminate the loop
if (j-i > k):
break
# Find the minimum value from i+1 to
# max (k or n)
if (arr[j] < arr[pos]):
pos = j
# Swap the elements from Minimum position
# we found till now to the i index
for j in range(pos, i, -1):
arr[j],arr[j-1] = arr[j-1], arr[j]
# Set the final value after swapping pos-i
# elements
k -= pos - i
# Driver Code
n, k = 5, 3
arr = [7, 6, 9, 2, 1]
minimizeWithKSwaps(arr, n, k)
# Print the final Array
for i in range(n):
print(arr[i], end = " ")C#
// C# program to find lexicographically
// minimum value after k swaps.
using System;
class GFG {
// Modifies arr[0..n-1] to lexicographically
// smallest with k swaps.
static void minimizeWithKSwaps(int []arr, int n,
int k)
{
for (int i = 0; i < n-1 && k > 0; ++i)
{
// Set the position where we want
// to put the smallest integer
int pos = i;
for (int j = i+1; j < n ; ++j)
{
// If we exceed the Max swaps
// then terminate the loop
if (j - i > k)
break;
// Find the minimum value from
// i + 1 to max k or n
if (arr[j] < arr[pos])
pos = j;
}
// Swap the elements from Minimum position
// we found till now to the i index
int temp;
for (int j = pos; j>i; --j)
{
temp=arr[j];
arr[j]=arr[j-1];
arr[j-1]=temp;
}
// Set the final value after
// swapping pos-i elements
k -= pos-i;
}
}
// Driver method
public static void Main()
{
int []arr = {7, 6, 9, 2, 1};
int n = arr.Length;
int k = 3;
// Function calling
minimizeWithKSwaps(arr, n, k);
// Print the final Array
for (int i=0; iphp
0; ++$i)
{
// Set the position where we want
// to put the smallest integer
$pos = $i;
for ($j = $i+1; $j < $n ; ++$j)
{
// If we exceed the Max swaps
// then terminate the loop
if ($j-$i > $k)
break;
// Find the minimum value from
// i+1 to max k or n
if ($arr[$j] < $arr[$pos])
$pos = $j;
}
// Swap the elements from Minimum
// position we found till now to
// the i index
for ($j = $pos; $j > $i; --$j)
{
$temp = $arr[$j];
$arr[$j] = $arr[$j-1];
$arr[$j-1] = $temp;
}
// Set the final value after
// swapping pos-i elements
$k -= $pos-$i;
}
//Print the final Array
for ($i = 0; $i < $n; ++$i)
echo $arr[$i] . " ";
}
// Driver code
$arr = array(7, 6, 9, 2, 1);
$n = count($arr);
$k = 3;
minimizeWithKSwaps($arr, $n, $k);
// This code is contributed by Sam007
?>Output: 2 7 6 9 1
时间复杂度： O(N 2 )
辅助空间： O(1)
参考：
http://stackoverflow.com/questions/25539423/finding-minimal-lexicographical-array