交换最大数量
在这个问题中，给出了一个正整数字符串，我们必须通过将数字的k次交换到不同的位置来找到其值最大的排列。
我们将通过选择一个数字并一次交换一个数字来解决此问题，以找到最大数字。我们重复该过程k次。回溯策略在这里起作用，因为当我们发现一个不大于先前值的数字时，我们便回溯到旧值并再次检查。
输入输出
Input: A number of multiple digits. The input is: 129814999 Output: The maximum value from these digits by swapping them. The output is: 999984211
算法
maxNum(number, swaps, maxNumber)
输入- 作为字符串的数字，交换数和maxNumber字符串。
输出-更新maxNumber以获取最大值。
Begin
if swaps = 0, then
return
n := number of digits in the number
for i := 0 to n-2, do
for j := i+1 to n-1, do
if number[i] < number[j], then
exchange number[i] and number[j]
if number is greater than maxNumber, then
maxNumber := number
maxNum(number, swaps-1, maxNumber)
exchange number[i] and number[j] again for backtrack
done
done
End示例
#include <iostream>
using namespace std;
void maxNum(string str, int swaps, string &max) {
if(swaps == 0) //when no swaps are left
return;
int n = str.length();
for (int i = 0; i < n - 1; i++) { //for every digits og given number
for (int j = i + 1; j < n; j++) {
if (str[i] < str[j]) { //when ith number smaller than jth number
swap(str[i], str[j]);
if (str.compare(max) > 0) //when current number is greater, make it maximum
max = str;
maxNum(str, swaps - 1, max); //go for next swaps
swap(str[i], str[j]); //when it fails, reverse the swapping
}
}
}
}
int main() {
string str = "129814999";
int swapNumber = 4;
string max = str;
maxNum(str, swapNumber, max);
cout <<"The given number is: " <<str << endl;
cout <<"The maximum number is: "<< max << endl;
}输出结果
The given number is: 129814999 The maximum number is: 999984211