153 & 154. Find Minimum in Rotated Sorted Array I & II

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.

Example 1:

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

Example 2:

Input: nums = [4,5,6,7,0,1,2]
Output: 0
Explanation: The original array was [0,1,2,4,5,6,7] and it was rotated 4 times.

Example 3:

Input: nums = [11,13,15,17]
Output: 11
Explanation: The original array was [11,13,15,17] and it was rotated 4 times. 

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.

My Solutions:

public int findMin(int[] nums) {
    int l = 0, r = nums.length - 1;
    while (l + 1 < r) {
        int mid = l + (r - l) / 2;
        if (nums[mid] > nums[r]) l = mid;
        else r = mid;
    }
    return nums[l] > nums[r] ? nums[r] : nums[l];
}

154. Find Mininum in Rotated Sorted Array II

和#153类似，但是会有重复数字。

My Solutions:

public int findMin(int[] nums) {
    int l = 0, r = nums.length - 1;
    while (l + 1 < r) {
        int mid = l + (r - l) / 2;
        if (nums[mid] == nums[r]) r--; // 多加这一步。如果有重复数字，这样可以往前找
        else if (nums[mid] > nums[r]) l = mid;
        else r = mid;
    }
    if (nums[l] <= nums[r]) return nums[l];
    else return nums[r];
}