某支付公司笔试编程题

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/**  * 最后一题的编程题  * 判断数组A(小数组) 是 数组B(大数组)的子集 数组都是有序且是整型数组  */ public class OptimizeTest {     public static void main(String[] args) { //        method01();         boolean b = method02();         System.out.println(b);     }       private static boolean method02() {         int[] bigArr = {1, 2, 3, 4, 5, 6, 7, 8};         int[] smallArr = {5, 6, 7};         //技术总监优化的方案         //先排除掉 压根没有重叠的元素         //在根据个数进行排除         //最后没有排除项了，在一一比较         //能最大限度的提高效率         if(bigArr[0] > smallArr[smallArr.length - 1]){             return false;         }else if(smallArr[0] > bigArr[bigArr.length - 1]){             return false;         }else{ //            int[] newBigArr = null;             int start = 0;             int end = 0;             for (int i = 0;i < bigArr.length;i++){                 if(bigArr[i] > smallArr[0]){                     start = i - 1;                     break;                 }             }             for (int i = bigArr.length - 1;i > 0 ;i--){                 if(bigArr[i] < smallArr[smallArr.length - 1]){                     end = i + 1;                     break;                 }             }             System.out.println(start);             System.out.println(end);             int len = end - start + 1;             if(len != smallArr.length){                 return false;             }else{                 bigArr = subArr(bigArr, start, end);             }             for (int i = 0;i < bigArr.length;i++){                 int m = 0;                 int k = i;                 for (int j = 0;j < smallArr.length;j++){                     if(bigArr[k] == smallArr[j]){                         m++;                         k++;                     }                 }                 if(m == smallArr.length){                     return true;                 }             }         }         return false;     }       /**      * 截取数组      * @param bigArr      * @param start      * @param end      * @return      */     private static int[] subArr(int[] bigArr, int start, int end) {         int[] newBigArr = new int[end - start + 1];         int i = 0;         while(start <= end){             newBigArr[i++] = bigArr[start];             start ++;         }         return newBigArr;     }   　　     private static boolean method01() {         int[] bigArr = {1, 2, 3, 4, 5, 6, 7, 8};         int[] smallArr = {5, 6, 7};         //我面试时的方案         for (int i = 0;i < bigArr.length;i++){             int m = 0;             int k = i;             for (int j = 0;j < smallArr.length;j++){                 if(bigArr[k] == smallArr[j]){                     m++;                     k++;                 }             }             if(m == smallArr.length){                 return true;             }         }         return false;     } }

　　结论: 我的方案没有考虑到效率，只是单纯完成了题目，对于数组小而言是没有什么问题的，但是一旦数组十分庞大时，效率就很低了，遍历一个一个比较，想想都有点可怕.......后面总监提醒了之后我回去按照他的思路实现了一下，先排除 没有交际的情况，在排除个数不相等的情况，在进行比较，比较的次数降低了，自然效率就高了。

以上个人测试，如有更好的方案，欢迎留言。