153 Find Minimum in Rotated Sorted Array

Leetcode

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.

边界条件，数组是顺序的，此时第一个元素小于最后一个，此时最小值是第一个元素。排除边界条件后，最小值是右半部分的第一个元素。使用二分法，当中间值小于第一个元素，右指针移到中间值，否则左指针移动到中间值右边，直到左右指针相遇。时间复杂度 o(log n)

public int findMin(int[] nums) {
    int first=nums[0];
    int right=nums.length;
    if(first<=nums[right-1]) return first;
    int left=0;
    int mid;
    while(left<right){
        mid=(left+right)/2;
        if(nums[mid]<first){
            right=mid;
        } else {
            left=mid+1;
        }
    }
    return nums[left];
}