1.按要求编写字符界面
- 编写一个字符界面的Java Application 程序,接受用户输入的10个整数,并输出这10个整数的最大值和最小值。
import java.util.Scanner;
public class MaxMin {
public static void main(String[] args) {
Arrays array = new Arrays();
array.setArr();
int max=array.getMax();
int min=array.getMin();
System.out.println("数组中最大值="+max);
System.out.println("数组中最小值="+min);
}
}
class Arrays {
private int[] arr;
public Arrays() {
arr = new int[10] ;
for(int i = 0; i<arr.length;i++) {
arr[i] =0;
}
}
public void setArr() {
Scanner sc = new Scanner(System.in);
System.out.println("请输入数组元素的值:");
arr = new int[10] ;
for(int i = 0; i<arr.length;i++) {
arr[i] = sc.nextInt();
}
}
public int getMax() {
int max = arr[0];
for(int i : arr) {
if(max < i)
max = i;
}
return max;
}
public int getMin() {
int min= arr[0];
for(int i : arr) {
if(min > i)
min= i;
}
return min;
}
}
2.最短回文串
- 给定一个字符串
s
,你可以通过在字符串前面添加字符将其转换为回文串。找到并返回可以用这种方式转换的最短回文串。
示例 1:
输入:s = "aacecaaa"
输出:"aaacecaaa"
示例 2:
输入:s = "abcd"
输出:"dcbabcd"
class Solution {
public static String shortestPalindrome(String s) {
StringBuilder r = new StringBuilder(s).reverse();
String str = s + "#" + r;
int[] next = next(str);
String prefix = r.substring(0, r.length() - next[str.length()]);
return prefix + s;
}
private static int[] next(String P) {
int[] next = new int[P.length() + 1];
next[0] = -1;
int k = -1;
int i = 1;
while (i < next.length) {
if (k == -1 || P.charAt(k) == P.charAt(i - 1)) {
next[i++] = ++k;
} else {
k = next[k];
}
}
return next;
}
}
3.交错字符串
给定三个字符串 s1
、s2
、s3
,请你帮忙验证 s3
是否是由 s1
和 s2
交错 组成的。
两个字符串 s
和 t
交错 的定义与过程如下,其中每个字符串都会被分割成若干 非空 子字符串:
s = s1 + s2 + ... + sn
t = t1 + t2 + ... + tm
|n - m| <= 1
交错 是 s1 + t1 + s2 + t2 + s3 + t3 + ... 或者 t1 + s1 + t2 + s2 + t3 + s3 + ...
提示:a + b
意味着字符串 a
和 b
连接。
示例 1:
输入:s1 = "aabcc", s2 = "dbbca", s3 = "aadbbcbcac"
输出:true
示例 2:
输入:s1 = "aabcc", s2 = "dbbca", s3 = "aadbbbaccc"
输出:false
示例 3:
输入:s1 = "", s2 = "", s3 = ""
输出:true
提示:
0 <= s1.length, s2.length <= 100
0 <= s3.length <= 200
s1、s2、和 s3 都由小写英文字母组成
class Solution {
public static void main(String[] args) {
String s1 = "aabcc";
String s2 = "dbbca";
String s3 = "aadbbcbcac";
Solution solu = new Solution();
boolean bool = solu.isInterleave(s1, s2, s3);
System.out.println(bool);
}
public boolean isInterleave(String s1, String s2, String s3) {
if ((s1.length() + s2.length()) != s3.length())
return false;
boolean[][] dp = new boolean[s2.length() + 1][s1.length() + 1];
dp[0][0] = true;
for (int i = 1; i <= s1.length(); i++) {
dp[0][i] = dp[0][i - 1] && s1.charAt(i - 1) == s3.charAt(i - 1) ? true : false;
}
for (int i = 1; i <= s2.length(); i++) {
dp[i][0] = dp[i - 1][0] && s2.charAt(i - 1) == s3.charAt(i - 1) ? true : false;
}
for (int i = 1; i < dp.length; i++) {
for (int j = 1; j < dp[0].length; j++) {
dp[i][j] = (dp[i][j - 1] && s1.charAt(j - 1) == s3.charAt(i + j - 1))
|| (dp[i - 1][j] && s2.charAt(i - 1) == s3.charAt(i + j - 1));
}
}
return dp[s2.length()][s1.length()];
}
}