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
    return findminNumberInRotateArray(nums, 0, numsLen - 1);
}
// 1 2 3 4 5
// 3 4 5 1 2


// 5 1 2 5 5 5 5 5 5
// 0 1 2 3 4 5 6 7 8
// left = 0, right = 8
// mid = 4
int findminNumberInRotateArray(int* nums, int left, int right) {
    if(left == right) return nums[left];
    int mid = left + (right - left) / 2;
    if(nums[mid] > nums[right]) {
        return findminNumberInRotateArray(nums, mid + 1, right);
    } else if(nums[mid] == nums[right]) {
        return findminNumberInRotateArray(nums, left, right - 1);
        // 4 4 4 3 4
        // 4 3 4 4 4
    }
        // 5 5 1 2 3
    return findminNumberInRotateArray(nums, left, mid);
}