算法-------二分查找

题目：https://leetcode.com/problems/search-in-rotated-sorted-array/

思路：
题目要求时间复杂度必须是 O(log n) 级别，所以用了二分查找
首先是使用二分查找，查找出数组中的转折点下标position，然后在position处对数组进行二分查找，查找出target的下标

解题：

public int search(int[] nums, int target) {
        if (nums == null || nums.length == 0) {
            return -1;
        }
        if (nums.length == 1) {
            if (nums[0] == target) {
                return 0;
            } else {
                return -1;
            }
        }
        //二分法查找中间转折点的下标
        int postion = -1;//转折点下标
        int left = 0;
        int right = nums.length - 1;
        while (left <= right) {
            int mid = left + (right - left) / 2;
            if (mid != left && nums[mid] < nums[mid - 1]) {
                postion = mid - 1;
                break;
            }
            if (mid != right && nums[mid] > nums[mid + 1]) {
                postion = mid;
                break;
            }
            if (nums[mid] < nums[left]) {//转折点在左侧范围，将right值修改
                right = mid - 1;
            } else {//转折点在右侧范围，将left值修改
                left = mid + 1;
            }
        }

        //使用二分法查找数据，从转折点的位置开始分开
        if (postion == -1) {//数组中没有转折点，数组是升序
            postion = findIndex(nums, 0, nums.length - 1, target);
        } else {//有转折点
            if (nums[postion] == target) {
                return postion;
            }
            if (target < nums[0]) {
                postion = findIndex(nums, postion + 1, nums.length - 1, target);
            } else {
                postion = findIndex(nums, 0, postion - 1, target);
            }
        }
        return postion;
    }

    /**
     * 从升序数组中找到target的index
     *
     * @param nums
     * @param target
     * @return
     */
    public int findIndex(int[] nums, int left, int right, int target) {
        while (left <= right) {
            int mid = left + (right - left) / 2;
            if (nums[mid] == target) {
                return mid;
            }
            if (nums[mid] > target) {
                right = mid - 1;
            } else {
                left = mid + 1;
            }
        }
        return -1;
    }