2012.9.19华为机试java标题

2012.9.19华为机试java题目
1、求数组里比平均数大的数的个数。
   int avgNum(int arg[])

package org.hhu.ztc;public class Test1 {public static void main(String args[]) {System.out.println(new Test1().avgNum(new int[] { 1, 2, 3, 4, 5, 6, 7,8, 9, 10 }));}public int avgNum(int s[]) {int sum = 0;int len = s.length;int flag = 0;double avg;for (int i = 0; i < len; i++) {sum = sum + s[i];}avg = (sum + 0.0) / len;for (int i = 0; i < len; i++) {if (s[i] >= avg)flag++;}return flag;}}

2、两大数相加。
用java.math.BigInteger类
注：必须是正整数相加

package org.hhu.ztc;import java.math.BigInteger;public class Test2 {public static void main(String args[]){System.out.println(new Test2().plusBigNum("123456789","987654321"));}public String plusBigNum(String arg1, String arg2){BigInteger big1 = new BigInteger(arg1);BigInteger big2 = new BigInteger(arg2);return big1.add(big2).toString();}}

不用java.math.BigInteger类

package org.hhu.ztc;public class BigIntegerOperation {public static void main(String args[]) {String num1 = "12345";String num2 = "54321";int len1 = num1.length();int len2 = num2.length();/** * 将不等长的两输入字符串格式化，比如 num1 = 123 num2 = 1234 经格式化后将成为num1 = 0123 num2 = * 1234为方便计算 */if (len1 > len2) {for (int i = 0; i < len1 - len2; i++) {num2 = "0" + num2;}} else if (len2 > len1) {for (int i = 0; i < len2 - len1; i++) {num1 = "0" + num1;}}int[] arr1 = BigIntegerOperation.str2intArr(num1);int[] arr2 = BigIntegerOperation.str2intArr(num2);int[] result = BigIntegerOperation.add(arr1, arr2);for (int i = result.length - 1; i >= 0; i--) {if (i == result.length - 1 && result[i] == 0)continue;System.out.print(result[i]);}}/** * 将字符串转换成整型数组 */public static int[] str2intArr(String str) {int len = str.length();int[] arr = new int[len];for (int i = 0; i < len; i++) {arr[i] = str.charAt(i) - '0';}return arr;}/** * 核心方法 两个整型数组相加 */public static int[] add(int a[], int b[]) {int maxLen = a.length;int[] sum = new int[maxLen + 1];for (int i = 0; i < maxLen; i++) {int tempSum = a[i] + b[i];sum[i] += tempSum % 10;int d = tempSum / 10; // 进位sum[i + 1] += d;}return sum;}}

3、求字符串中最大回文子串。
如输入ababbac 输出abba

package org.hhu.ztc;public class Test3 {public static void main(String args[]) {System.out.println(new Test3().getPalindromeString("ababbac"));}public String getPalindromeString(String inputString) {int len = inputString.length();for (int i = 0; i < len; i++) {for (int j = 0; j < i + 1; j++) {String sub = inputString.substring(j, len - i + j);String subReverse = new StringBuffer(sub).reverse().toString();if (sub.equals(subReverse))return sub;}}return "无回文子串";}}

3.补充，求出所有回文子串及最大回文子串

package org.hhu.ztc;public class Test3 {public static void main(String args[]) { new Test3().getPalindromeString("ababbac");}public String[] getPalindromeString(String inputString) {int len = inputString.length();String s1[] = new String[len * (len + 1) / 2];String s2[] = new String[len * (len + 1) / 2];int n = 0;for (int i = 0; i < len; i++) {for (int j = 0; j < i + 1; j++) {String sub = inputString.substring(j, len - i + j);String subReverse = new StringBuffer(sub).reverse().toString();if (sub.equals(subReverse)) {s1[n] = sub;n++;}}}int m = 0;int maxlen = 0;if (s1[0] != null) {maxlen = s1[0].length();for (int i = 0; i < n; i++) {if (s1[i].length() == maxlen) {s2[m] = s1[i];m++;}}}//未考虑重复System.out.println("所有回文子串：");for(int i=0;i<n;i++){System.out.println(s1[i]);}System.out.println("最大回文子串：");for(int i=0;i<m;i++){System.out.println(s2[i]);}return s2;}}

1 楼 wavechao 2012-11-12   感觉求回文子串的有难度

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