题意：给你一个二进制串，试问是否能够通过删除一些不相邻的字符，以使得二进制串变成非降序串。
思路：原理很简单，不用管前面的0，只要满足第一次出现相邻的“11”时，后面不会再出现相邻的“00”即可。
代码实现C++：
#include<bits/stdc++.h>
#define endl '\n'
#define null NULL
#define ll long long
#define int long long
#define pii pair<int, int>
#define lowbit(x) (x &(-x))
#define ls(x) x<<1
#define rs(x) (x<<1+1)
#define me(ar) memset(ar, 0, sizeof ar)
#define mem(ar,num) memset(ar, num, sizeof ar)
#define rp(i, n) for(int i = 0, i < n; i ++)
#define rep(i, a, n) for(int i = a; i <= n; i ++)
#define pre(i, n, a) for(int i = n; i >= a; i --)
#define IOS ios::sync_with_stdio(0); cin.tie(0);cout.tie(0);
const int way[4][2] = {
{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
using namespace std;
const int inf = 0x3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-6;
const ll mod = 1e9+7;
const int N = 2e5 + 5;
inline void read(int &x){
char t=getchar();
while(!isdigit(t)) t=getchar();
for(x=t^48,t=getchar();isdigit(t);t=getchar()) x=x*10+(t^48);
}
int t;
string s;
signed main()
{
IOS;
cin >> t;
while(t --){
cin >> s;
string s1 = s;
sort(s1.begin(), s1.end());
if(s==s1){
cout << "YES" << endl;
continue;
}
int flag = 0;
for(int i = 0; i < s.size(); i ++){
if(s[i]=='1'&&s[i+1]=='1'){
for(int j = i+2; j < s.size(); j ++){
if(s[j]=='0'&&s[j+1]=='0'){
flag = 1;
break;
}
}
if(flag) break;
}
}
cout << (flag ? "NO":"YES") << endl;
}
return 0;
}Java练习：
import java.util.Scanner;
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int t = in.nextInt();
while(t>0) {
t --;
String s = in.next();
int len = s.length(), flag = 0;
for(int i = 0; i < len-1; i ++){
if(s.charAt(i)=='1' && s.charAt(i+1)=='1'){
for(int j = i+2; j < len-1; j ++){
if(s.charAt(j)=='0' && s.charAt(j+1)=='0'){
flag = 1;
break;
}
}
}
if(flag!=0) break;
}
if(flag!=0) System.out.println("NO");
else System.out.println("YES");
}
in.close();
}
}