二分查找循环有序数组

二分法查找循环有序数组中的某一元素,如果存在返回该值，否则返回 -1。
int getNumFromLooOrderArray(int arr[], int target, int left, int right) {

    while (left < right) {
        int mid = left + (right - left) / 2;
        cout<<"left: " << left << " right: " << right << endl;
        if (arr[mid] == target) {
            return mid;
        }
        if (arr[mid] < arr[right]) {
             // mid 右边有序
            if (target > arr[mid] && target <= arr[right]) {
                left = mid + 1;
            } else {
                right = mid - 1;
            }
            //if (target > arr[right]) {
              //  right = mid - 1;
            //}
        } else{
             // mid 左边有序
            if (target < arr[mid] && target >= arr[left]) {
                right = mid - 1;
            } else {
                left = mid + 1;
            }
        }
    }
    if (target == arr[left]) {
        return left;
    }
    return -1;
}



int main()
{
    int dp[6]={4,5,6,1,2,3};
    cout  << getNumFromLooOrderArray( dp, 1,0,5 );


}

　

//使用二分法查找循环有序数组的最小值
int getSubScriptFromArray(int arr[], int left, int right){
    while (left < right) {
        int mid = left + (right - left) / 2;
        if (arr[mid] >arr[right]) {
            //左边有序， 最小值在右边，不包含mid
            left = mid + 1;
        } else if (arr[mid] < arr[right]) {
            // 右边有序， 最小值在左边，可能是mid
            right = mid;
        } else{
            right = right - 1;
        }
    }
    return arr[left];
}