题解 | #旋转数组的最小数字#

/**
 * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
 *
 * 
 * @param nums int整型一维数组 
 * @param numsLen int nums数组长度
 * @return int整型
 */
int minNumberInRotateArray(int* nums, int numsLen ) {
    // write code here
    int i = 0,j = 0;
    int low = 0;
    int high = numsLen-1;
    int mid = (low+high)/2;

    while (1) {
        if (high-low+1<3) {
            if(nums[low]>nums[low+1]) return nums[low+1];
            else if(nums[high-1]>nums[high]) return nums[high];
        }
        //if(nums[mid-1]>nums[mid]) return nums[mid];
        if (nums[mid]<nums[low])high = mid;
        else if (nums[mid]>nums[low]) low = mid;
        else if (nums[mid]=nums[low]) //考虑特例
        {
            for(i = 0; i<numsLen-1; i++)       //10111这种只有暴力循环了
            {
                if(nums[i]>nums[i+1]) return nums[i+1];
            }
            return nums[0];   ////0111这种不循环的
        }
        mid = (low+high)/2;
    }
    return nums[0];

}