ui设计app界面图片/网站优化 seo和sem

一. 题目

把一个数组最开始的若干个元素搬到数组的末尾,我们称之为数组的旋转.输入一个递增排序的数组的一个旋转,输出旋转数组的最小元素.例如数组{3,4,5,1,2}为{1,2,3,4,5}的一个旋转,该数组的最小值为1.

代码请到我的代码库中下载 Point2Offer

二. 代码

package ween_2;
/*** 剑指offer: 旋转数组的最小数字* 方法:二分查找+顺序查找针对特殊情况* 测试用例:功能测试(34512,34513,12345,1,10111)* 特殊输入测试(数组为空)* @author dingding* Date:2017-6-21 12:40* Declaration: All Rights Reserved!*/
public class No8 {public static void main(String[] args) {test1();test2();test3();test4();test5();test6();test7();}//处理private static int findMinNumberInArray(int[] arr,int length) throws NumberFormatException{if (arr == null || length<1) {throw new NumberFormatException("illegal number input."); }int pBegin = 0;int pEnd = length-1;int pMid = pBegin;  //处理类似12345情形while (arr[pBegin]>=arr[pEnd]){if (pEnd-pBegin == 1) {return arr[pEnd];}//取中间位置pMid = (pBegin+pEnd)/2;//两端和中间的数都相等,只能顺序处理if (arr[pMid]==arr[pBegin] && arr[pMid]==arr[pEnd]) {return minInOrder(arr,pBegin,pEnd);}if (arr[pMid]>=arr[pBegin]) {pBegin = pMid;}else if (arr[pEnd]>=arr[pMid]) {pEnd = pMid;}}return arr[pMid];}//顺序处理 10111类似情形,同时处理单字符private static int minInOrder(int[] arr, int pBegin, int pEnd) {int min = arr[pBegin];for (int i=pBegin+1;i<=pEnd;i++) {if (arr[i]<min) {min = arr[i];}}return min;}/*=====================测试用例==================*/private static void test(int[] arr){if (arr == null) {System.out.println("数组为空!");return;}int length = arr.length;int result = findMinNumberInArray(arr, length);System.out.println("结果为: "+result);System.out.println("============================");}private static void test1() {int[] arr = {3,4,5,1,2};test(arr);}private static void test2() {int[] arr = {3,4,5,1,1,2};test(arr);}private static void test3() {int[] arr = {3,4,5,1,2,2};test(arr);}private static void test4() {int[] arr = {1, 0, 1, 1, 1};test(arr);}private static void test5() {int[] arr = {1,2,3,4,5};test(arr);}private static void test6() {int[] arr = {7};test(arr);}private static void test7() {test(null);}}

有不妥当之处,麻烦告知:D