33. Search in Rotated Sorted Array[Medium]

33. Search in Rotated Sorted Array

There is an integer array nums sorted in ascending order (with distinct values).

Prior to being passed to your function, nums is possibly rotated at an unknown pivot index k (1 <= k < nums.length) such that the resulting array is [nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]] (0-indexed). For example, [0,1,2,4,5,6,7] might be rotated at pivot index 3 and become [4,5,6,7,0,1,2].

Given the array nums after the possible rotation and an integer target, return the index of target if it is in nums, or -1 if it is not in nums.

You must write an algorithm with O(log n) runtime complexity.

Constraints:

• 1 <= nums.length <= 5000
• -10^4 <= nums[i] <= 10^4
• All values of nums are unique.
• nums is an ascending array that is possibly rotated.
• -10^4 <= target <= 10^4

Example

Input: nums = [4,5,6,7,0,1,2], target = 0
Output: 4

思路

二分查找变体，在二分查找基础上，再多一步条件判断，看当前处于哪一段

题解

    public int search(int[] nums, int target) {
        int left, right, mid;
        left = 0;
        right = nums.length - 1;

        while (left <= right) {
            mid = (left + right) / 2;
            if (nums[mid] == target)
                return mid;
            // 如果满足，说明左半段是顺序的
            if (nums[mid] >= nums[left]) {
                // 那如果比最左小或者比中位大，那只可能在右半段了
                if (target < nums[left] || target > nums[mid])
                    left = mid + 1;
                else
                    right = mid - 1;
            } else {
                // 说明右半段是顺序的，刨除右半部分（比最右大或比中位小），那就是左半段
                if (target > nums[right] || target < nums[mid])
                    right = mid - 1;
                else
                    left = mid + 1;
            }

        }
        return -1;
    }