*153. Find Minimum in Rotated Sorted Array[Medium]

153. Find Minimum in Rotated Sorted Array

Suppose an array of length n sorted in ascending order is rotated between 1 and n times. For example, the array nums = [0,1,2,4,5,6,7] might become:

• [4,5,6,7,0,1,2] if it was rotated 4 times.
• [0,1,2,4,5,6,7] if it was rotated 7 times.

Notice that rotating an array [a[0], a[1], a[2], ..., a[n-1]] 1 time results in the array [a[n-1], a[0], a[1], a[2], ..., a[n-2]].

Given the sorted rotated array nums of unique elements, return the minimum element of this array.

You must write an algorithm that runs in O(log n) time.

Constraints:

• n == nums.length
• 1 <= n <= 5000
• -5000 <= nums[i] <= 5000
• All the integers of nums are unique.
• nums is sorted and rotated between 1 and n times.

Example

Input: nums = [3,4,5,1,2]
Output: 1
Explanation: The original array was [1,2,3,4,5] rotated 3 times.

思路

二分查找变体，虽然被旋转了，但子串还是有序的，通过中位来判断

题解

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

        while (left <= right) {
            // 如果当前最右比最左大，那说明这一段子串是顺序的，直接取最左
            if (nums[right] > nums[left]) {
                min = Math.min(min, nums[left]);
                break;
            }

            mid = (left + right) / 2;
            min = Math.min(nums[mid], min);

            // 如果当前中位比最右大，那说明右边这段应该是被旋转了，是属于小的那一部分，就切到右半边来
            if (nums[mid] > nums[right])
                left = mid + 1;
            else
                // 如果当前中位比最右小，那没有旋转，左边是小的部分，切到左半边
                right = mid - 1;

        }
        return min;
    }