通过在C ++中按一定顺序排列数字来获得最大数字
在这个问题中，我们得到了一个数字数组，我们必须找到可以通过以某种方式更改它们而获得的最大值。安排的条件是，偶数和奇数的顺序应保持相同，即所有偶数的顺序均不可更改。
让我们举个例子来更好地理解这个概念，
Input : {17, 80, 99, 27, 14 , 22}
Output: 801799271422
Explanation: the order of Even and Odd numbers is :
Even : 80 14 22
Odd : 17 99 27这里99是最大的数字，但是17以奇数的顺序排在前面，因此我们先考虑80，然后依次进行排列，例如-801799271422
既然我们已经理解了问题，那么我们就尝试为此解决方案。在这里，由于定义了关于偶数和奇数的顺序的约束，因此我们不能追求经典降序。因此，我们将必须保持此顺序，并检查“偶数”和“奇数”订单的前几个元素中的最大元素。然后像那样去让我们看一下可以使这一点更加清楚的算法。
算法
Step 1 : Create two structures, one for even other for odd, this will maintain the sequence.
Step 2 : Take one element from each structure and check
which combination makes a large number. Example, if E is
the even number and O is the odd number which are at the
top of the structure. then we will check which one is Greater of EO and OE.
Step 3 : Place the greater combination into the final sequence.
Step 4 : Print the final sequence.示例
现在，让我们基于该算法创建一个程序。
#include <bits/stdc++.h>
using namespace std;
string merge(vector<string> arr1, vector<string> arr2) {
int n1 = arr1.size();
int n2 = arr2.size();
int i = 0, j = 0;
string big = "";
while (i < n1 && j < n2) {
if ((arr1[i]+arr2[j]).compare((arr2[j]+arr1[i])) > 0)
big += arr1[i++];
else
big += arr2[j++];
}
while (i < n1)
big += arr1[i++];
while (j < n2)
big += arr2[j++] ;
return big;
}
string largestNumber(vector<string> arr, int n) {
vector<string> even, odd;
for (int i=0; i<n; i++) {
int lastDigit = arr[i].at(arr[i].size() - 1) - '0';
if (lastDigit % 2 == 0)
even.push_back(arr[i]);
else
odd.push_back(arr[i]);
}
string biggest = merge(even, odd);
return biggest;
}
int main() {
vector<string> arr;
arr.push_back("17");
arr.push_back("80");
arr.push_back("99");
arr.push_back("27");
arr.push_back("14");
arr.push_back("22");
int n = arr.size();
cout<<"Biggest possible number from the array is = "<<largestNumber(arr, n);
return 0;
}输出结果
Biggest possible number from the array is = 801799271422