字节笔试 后端笔试题（8.21）
总体来说，比较复杂，做一题老半天，真心觉得两个小时也就最多做三道题。
8.22更新：问了一下字节hr，挂了。
做题情况：100 100 75 1.6 下面附前三题的代码，第四题没时间看了（顺便有没有大佬帮忙看看我第三题的忘记考虑哪个约束了吗）下面附代码
1、10的数学规律，无非就是找1的个数，然后判断有没有1的位置是不是在最后或者最前面
假设1的个数是n
如果 1在第一个位置也在最后一个位置，那么 和最小=n*11-10-1
如果 1在第一个位置但不在最后一个位置，那么 和最小=n*11-1
如果 1不在第一个位置但是在最后一个位置，那么 和=n*11-10
如果 1不在第一个位置也不在最后一位置，那么 和=n*11
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int t = Integer.valueOf(sc.nextLine());
for (int i = 0; i <t ; i++) {
String[] strings = sc.nextLine().split(" ");
int n = Integer.valueOf(strings[0]);
int k = Integer.valueOf(strings[1]);
String s =sc.nextLine();
int oneNum = 0;
for (int j = 0; j < s.length(); j++) {
if(s.charAt(j)=='1'){
oneNum++;
}
}
if(oneNum==0){
System.out.println(0);
}else if(oneNum==1){
if(k>=1){
int index = s.length()-1;
for (int j = s.length()-1; j >=0; j--) {
if(s.charAt(j)=='1'){
index=j;
break;
}
}
if(k>=s.length()-1-index){
System.out.println(1);
}else if(k<s.length()-1-index&&k>=index){
System.out.println(10);
}else {
System.out.println(11);
}
}else {
if(s.charAt(n-1)=='1'){
System.out.println(1);
}else if(s.charAt(0)=='1'&&s.length()==1){
System.out.println(1);
}else if(s.charAt(0)=='1'&&s.length()>1){
System.out.println(10);
}else {
System.out.println(11);
}
}
}else{
if(k==0){
if(s.charAt(n-1)=='1'&&s.charAt(0)!='1'){
System.out.println(oneNum*11-10);
}else if(s.charAt(0)=='1'&&s.charAt(n-1)=='1'){
System.out.println(oneNum*11-10-1);
}else if(s.charAt(0)=='1'&&s.charAt(n-1)!='1'){
System.out.println(oneNum*11-1);
}else {
System.out.println(oneNum*11);
}
}else {
int r = s.length()-1;
for (int j = s.length()-1; j >=0; j--) {
if(s.charAt(j)=='1'){
r=j;
break;
}
}
int l = s.length()-1;
for (int j = 0; j <n; j++) {
if(s.charAt(j)=='1'){
l=j;
break;
}
}
if(k>=s.length()-1-r){
int remain = k-s.length()+1+r;
if(remain>=l){
System.out.println(oneNum*11-10-1);
}else{
System.out.println(oneNum*11-10);
}
}else{
if(k>=l){
System.out.println(oneNum*11-1);
}else {
System.out.println(oneNum*11);
}
}
}
}
}
} 2、O出口 就反向思维，从O开始用BFS找能到达的地方就行了，最后输出M*N-能到的地方的个数就是不能到达O出口的地方个数
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
String[] strings = sc.nextLine().split(" ");
int m = Integer.valueOf(strings[0]);
int n = Integer.valueOf(strings[1]);
char[][] grid = new char[m][n];
for (int i = 0; i < m; i++) {
String s = sc.nextLine();
for (int j = 0; j < n; j++) {
grid[i][j] = s.charAt(j);
}
}
Deque<int[]> deque = new LinkedList<>();
boolean[][] visited = new boolean[m][n];
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if(grid[i][j]=='O'){
deque.addLast(new int[]{i,j});
visited[i][j]=true;
break;
}
}
}
int count = 1;
while (!deque.isEmpty()){
int[] start = deque.pollFirst();
int x =start[0];
int y =start[1];
if(x-1>=0&&(grid[x-1][y]=='.'||grid[x-1][y]=='D')&&!visited[x-1][y]){
deque.addLast(new int[]{x-1,y});
visited[x-1][y]=true;
count++;
}
if(x+1<m&&(grid[x+1][y]=='.'||grid[x+1][y]=='U')&&!visited[x+1][y]){
deque.addLast(new int[]{x+1,y});
visited[x+1][y]=true;
count++;
}
if(y-1>=0&&(grid[x][y-1]=='.'||grid[x][y-1]=='R')&&!visited[x][y-1]){
deque.addLast(new int[]{x,y-1});
visited[x][y-1]=true;
count++;
}
if(y+1<n&&(grid[x][y+1]=='.'||grid[x][y+1]=='L')&&!visited[x][y+1]){
deque.addLast(new int[]{x,y+1});
visited[x][y+1]=true;
count++;
}
}
System.out.println(m*n-count);
} 3.{}匹配，只过了70%。。。还请各位大佬帮忙指点一下漏判了哪些条件 public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = Integer.valueOf(sc.nextLine());
String t = sc.nextLine();
List<String> list = new ArrayList<>();
List<Integer> l = new ArrayList<>();
List<Integer> r = new ArrayList<>();
for (int i = 0; i < t.length(); i++) {
if(t.charAt(i)=='{'){
l.add(i);
}
if(t.charAt(i)=='}'){
r.add(i);
}
}
if(l.size()!=0){
String sub = t.substring(0,l.get(0));
if(sub!=null&&sub.length()!=0&&sub!="")list.add(sub);
for (int i = 0; i < r.size()-1; i++) {
String substr = t.substring(r.get(i)+1,l.get(i+1));
if(substr!=null&&substr.length()!=0&&substr!="") list.add(substr);
}
String sunlast = t.substring(r.get(r.size()-1)+1);
if(sunlast!=""&&sunlast!=null&&sunlast.length()!=0){
list.add(sunlast);
}
}else {
list.add(t);
}
for (int i = 0; i < n; i++) {
String s = sc.nextLine();
boolean flag = true;
if(s.charAt(0)=='{'&&s.charAt(s.length()-1)=='}'&&list.size()==0) {
System.out.println("False");
continue;
}
if(t.charAt(0)!='{'&&t.charAt(t.length()-1)!='}'){
String first = list.get(0);
String last = list.get(list.size()-1);
if(!s.substring(0,first.length()).equals(first)||!s.substring(s.length()-last.length(),s.length()).equals(last)){
System.out.println("False");
continue;
}else{
String sub = "";
for (int j = 0; j < list.size(); j++) {
sub = list.get(j);
int index = s.indexOf(sub);
if(index==-1){
System.out.println("False");
flag=false;
}else{
sub = s.substring(index+sub.length());
}
}
if(flag==false) continue;
}
}else if(t.charAt(0)=='{'&&t.charAt(t.length()-1)!='}'){
String last = list.get(list.size()-1);
if(!s.substring(s.length()-last.length(),s.length()).equals(last)){
System.out.println("False");
continue;
}else{
String sub = "";
for (int j = 0; j < list.size(); j++) {
sub = list.get(j);
int index = s.indexOf(sub);
if(index==-1){
System.out.println("False");
flag=false;
}else{
sub = s.substring(index+sub.length());
}
}
if(flag==false) continue;
}
}else if(t.charAt(0)!='{'&&t.charAt(t.length()-1)=='}'){
String first = list.get(0);
if(!s.substring(0,first.length()).equals(first)){
System.out.println("False");
continue;
}else{
String sub = "";
for (int j = 0; j < list.size(); j++) {
sub = list.get(j);
int index = s.indexOf(sub);
if(index==-1){
System.out.println("False");
flag=false;
}else{
sub = s.substring(index+sub.length());
}
}
if(flag==false) continue;
}
}else{
String sub = "";
for (int j = 0; j < list.size(); j++) {
sub = list.get(j);
int index = s.indexOf(sub);
if(index==-1){
System.out.println("False");
flag=false;
}else{
sub = s.substring(index+sub.length());
}
}
if(flag==false) continue;
}
System.out.println("True");
}
} 4.概率题，偷了一点分，emmmm
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