Find Minimum in Rotated Sorted Array
描述
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.
分析
从左向右扫描,扫描到的第一个逆序的位置,肯定是原始数组中第一个元素,时间复杂度O(n)
。
不过本题依旧可以用二分查找,最关键的是要判断那个“断层”是在左边还是右边。
- 若
A[mid] < A[right]
,则区间[mid,right]
一定递增,断层一定在左边 - 若
A[mid] > A[right]
,则区间[left,mid]
一定递增,断层一定在右边 nums[mid] == nums[right]
,这种情况不可能发生,因为数组是严格单调递增的,不存在重复元素
代码
// Find Minimum in Rotated Sorted Array
// 时间复杂度O(logn),空间复杂度O(1)
public class Solution {
public int findMin(int[] nums) {
int left = 0;
int right = nums.length - 1;
while (left < right) {
int mid = left + (right - left) / 2;
if (nums[mid] < nums[right]) {
right = mid;
} else {
left = mid + 1;
}
}
return nums[left];
}
}