求旋转数组中的最小值

packagecom.ytx.Search;

/**

 * 在旋转数组中找到最小的数字

 * 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。

 * {1,2,3,4,5}-> {3,4,5,1,2}

 * 特例：

   把排序数组的前面的0个元素搬到最后面，即排序数组本身。

 * 还要考虑多个重复数字的情况如：

 * {0,1,1,1,1,}-> {1,0,1,1,1}或{1,1,1,0,1}

 *@authoryuantian xin

 *

 */

publicclass SearchMin_inRotateArray {

      publicint findMin(int[]nums) {

            intlen = nums.length;

            if(nums== null || len <= 0) {

                  System.out.println("数组未初始化");

                  System.exit(-1);

            }

            intlow = 0;

            inthigh = len - 1;

            intmid = low;

            while(nums[low] >= nums[high]) {

                  if((high- low) == 1) {

                        mid= high;

                        break;

                  }

                  mid= (low+ high) /2;

                  //如果高指针，低指针和中间指针指向的数字相等，直接顺序查找

                  if(nums[low]==nums[high] && nums[low]==nums[mid]) {

                        returnorderSearch(nums,low,high);

                  }

                  

                  if(nums[mid] >= nums[low]) {

                        low= mid;

                  }elseif(nums[mid] <=nums[high]){

                        high= mid;

                  }

            }

            returnnums[mid];

      }

      //直接顺序查找

      publicint orderSearch(int[]nums,int low, int high) {

            intmin = nums[low];

            for(inti = low;i <=high;i++) {

                  if(min> nums[i])

                        min= nums[i];

            }

            returnmin;

      }

      publicstatic void main(String[]args) {

            int[]data = {1,1,1,0,1};

            SearchMin_inRotateArrayob = newSearchMin_inRotateArray();

            intminNumber = ob.findMin(data);

            System.out.println(minNumber);

      }

}