Suppose a sorted array is rotated at some pivot unknown to you beforehand.
(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).
Find the minimum element.
You may assume no duplicate exists in the array.
数列基本有序，则使用二分查找。假设数列是n，s是起始下标，e是最后下标，m是中间元素下标。分为三种情况：
ns < ne
nm > ns > ne
nm < ne < ns
情况1：ns < ne
0 1 2 4 5 6 7
这种情况，直接返回ns。因为数列是有序的，或者只是旋转过一次。如果ns < ne，则表明没有旋转。最小元素是ns。
情况2：nm > ns > ne
代码语言：txt复制4 5 6 7 0 1 2
public int findMin(int[] nums) {
if (nums.length == 1) {
return nums[0];
}
if (nums.length == 2) {
return Math.min(nums[0], nums[1]);
}
int s = 0, e = nums.length - 1;
int m = (s e) / 2;
if (nums[s] < nums[e]) {
return nums[s];
}
if (nums[m] > nums[s]) {
return findMin(Arrays.copyOfRange(nums, m 1, e 1));
}
return findMin(Arrays.copyOfRange(nums, s, m 1));
}
