面试题 38: 数字在排序数组中出现的次数

一. 题目

统计一个数字在排序数组中出现的次数.例如输入排序数组{1,2,3,3,3,3,4,5} 和数字3,由于3在这个数组中出现了4次,因此输出4.

代码请到我的代码库中下载 Point2Offer

二. 代码

package ween_2;/**难度系数:*** * 剑指offer: 数字在排序数组中出现的次数 * 方法:二分查找 * 测试用例:功能测试(不存在,存在1次,多次,最大值,最小值) * 特殊输入测试(数组为空,数组中只有一个数字) * @author dingding * Date:2017-6-27 20:00 * Declaration: All Rights Reserved! */public class No38 {    public static void main(String[] args) {        test1();        test2();        test3();        test4();        test5();        test6();        test7();    }    //返回第一个k的下标    private static int getFirstK(int[] array,int length,int k,int start,int end){        if (start>end) {            return -1;        }        int middleIndex = (start+end)/2;        int middleData = array[middleIndex];        if (middleData == k) {            if ((middleIndex > 0 && array[middleIndex-1]!=k) || middleIndex==0) {                return middleIndex;            }else {                end = middleIndex-1;            }        }else if (middleData>k) {            end = middleIndex - 1;        }else {            start = middleIndex + 1;        }        return getFirstK(array, length, k, start, end);    }    //返回最后一个k的下标    private static int getLastK(int[] array,int length,int k,int start,int end){        if (start>end) {            return -1;        }        int middleIndex = (start+end)/2;        int middleData = array[middleIndex];        if (middleData==k) {            if ((middleIndex<length-1 && array[middleIndex+1]!=k) || middleIndex==length-1) {                return middleIndex;            }else {                start = middleIndex+1;            }        }else if(middleData<k){            start = middleIndex + 1;        }else {            end = middleIndex - 1;        }        return getLastK(array, length, k, start, end);    }    //计算出现的次数    private static int getNumberOfK(int[] array,int length,int k){        int number = 0;        if (array == null || length<1) {            return -1;        }        if (array != null && length>0) {            int first = getFirstK(array, length, k, 0, length-1);            int last = getLastK(array, length, k, 0, length-1);            if (first>-1 && last > -1) {                number = last - first +1;            }        }        return number;    }    /*==================测试用例================*/    private static void test(int[] array,int length,int k){        int result = getNumberOfK(array, length, k);        if (result != -1) {            System.out.println(k+"出现的次数为: "+ result);        }else{            System.out.println("不是有效输入!");        }        System.out.println("====================");    }    private static void test1() {        int[] array = {1,2,3,4,5,5,6};        int k=7;        test(array, array.length, k);    }    private static void test2() {        int[] array = {1,2,3,4,5,5,6};        int k=3;        test(array, array.length, k);    }    private static void test3() {        int[] array = {1,2,3,4,5,5,6};        int k=5;        test(array, array.length, k);    }    private static void test4() {        int[] array = {1,2,3,4,5,5,6};        int k=6;        test(array, array.length, k);    }    private static void test5() {        int[] array = {1,2,3,4,5,5,6};        int k=1;        test(array, array.length, k);           }    private static void test6() {        int[] array = {1};        int k=1;        test(array, array.length, k);           }    private static void test7() {        int[] array = {};        int k=6;        test(array, array.length, k);    }}

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有不妥当之处,麻烦告知:D