Given a non-negative integer, you could swap two digits at most once to get the maximum valued number. Return the maximum valued number you could get.

Example 1:

Input: 2736
Output: 7236
Explanation: Swap the number 2 and the number 7.

Example 2:

Input: 9973
Output: 9973
Explanation: No swap.

Note:

  1. The given number is in the range [0, 108]

思路:

暴力枚举所有交换可能,竟然没有超时。。。感觉因为输入最多到108

最多8位,复杂度O(n3)虽然很高,但数据规模小,但总之不是什么好办法。

vector<int> digits(int num)
{
if (num == )return{};
vector<int> ret;
while (num > )
{
ret.push_back(num % );
num /= ;
}
reverse(ret.begin(), ret.end());
return ret;
}
int toNum(vector<int>num)
{
int r = ;
for (int i = ; i < num.size();i++)
{
r *= ;
r += num[i];
}
return r;
}
int maximumSwap(int n)
{
vector<int> num = digits(n);
int m = n; for (int i = ; i < num.size();i++)
{
for (int j = i + ; j < num.size();j++)
{
swap(num[i], num[j]);
m = max(m, toNum(num));
swap(num[i], num[j]);
}
}
return m;
}
 

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