Minimize The Integer CodeForces - 1251C（思维）

You are given a huge integer a consisting of n digits (n is between 1 and 3⋅105, inclusive). It may contain leading zeros.

You can swap two digits on adjacent (neighboring) positions if the swapping digits are of different parity (that is, they have different remainders when divided by 2).

For example, if a=032867235 you can get the following integers in a single operation:

302867235 if you swap the first and the second digits;
023867235 if you swap the second and the third digits;
032876235 if you swap the fifth and the sixth digits;
032862735 if you swap the sixth and the seventh digits;
032867325 if you swap the seventh and the eighth digits.
Note, that you can’t swap digits on positions 2 and 4 because the positions are not adjacent. Also, you can’t swap digits on positions 3 and 4 because the digits have the same parity.

You can perform any number (possibly, zero) of such operations.

Find the minimum integer you can obtain.

Note that the resulting integer also may contain leading zeros.

Input
The first line contains one integer t (1≤t≤104) — the number of test cases in the input.

The only line of each test case contains the integer a, its length n is between 1 and 3⋅105, inclusive.

It is guaranteed that the sum of all values n does not exceed 3⋅105.

Output
For each test case print line — the minimum integer you can obtain.

Example
Input
3
0709
1337
246432
Output
0079
1337
234642
Note
In the first test case, you can perform the following sequence of operations (the pair of swapped digits is highlighted): 070−−9→0079.

In the second test case, the initial integer is optimal.

In the third test case you can perform the following sequence of operations: 24643−−2→2463−−42→243−−642→234642.
这个题目根codeforces上另一个题目思路很像，或者说这种题的思路都很相像。
由于只能奇偶不同的时候交换，因此奇数的相对位置和偶数的相对位置是不会改变的。因此我们把奇数找出来，偶数找出来，每一次都输出相对小的那个数就可以了。这个思路很常用。
代码如下：

#include<bits/stdc++.h>
#define ll long long
using namespace std;

const int maxx=3e5+100;
int s1[maxx];
int s2[maxx];
string s;
int n;

int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		cin>>s;
		n=s.length();
		int len1=0,len2=0;
		for(int i=0;i<n;i++)
		{
			if((s[i]-'0')&1) s1[len1++]=s[i]-'0';
			else s2[len2++]=s[i]-'0';
		}
		int i=0,j=0;
		while(i<len1&&j<len2)
		{
			if(s1[i]<s2[j]) cout<<s1[i++];
			else cout<<s2[j++];
		}
		if(i==len1) while(j<len2) cout<<s2[j++];
		else while(i<len1) cout<<s1[i++];
		cout<<endl;
	}
	return 0;
}

努力加油a啊，(^o)/~