154. Find Minimum in Rotated Sorted Array II
Follow up for "Find Minimum in Rotated Sorted Array":
What if duplicates are allowed?Would this affect the run-time complexity? How and why?
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.
The array may contain duplicates.
153的扩展。二分查找,如果mid小于最右侧,证明右边是sort好的,则最小值必定在左边。
如果mid大于最右侧,则证明左侧最小值被rotate到了右侧,则最小值必定在右侧。
如果中间值等于右侧值,则无法判断哪边是sort好的,比如1,2,3,3,3,3,也可以是3,3,3,3,1,2,3。因此需要一个个的收缩,最坏情况下时间复杂度为O(n).
public class Solution {
public int findMin(int[] num) {
int min = num[0];
int left = 0;
int right = num.length - 1;
while (left <= right) {
int mid = left + (right - left) / 2;
if (num[mid] < num[right]) {
right = mid - 1;
} else if (num[mid] > num[right]) {
left = mid + 1;
} else {
right --;
}
min = Math.min(min, num[mid]);
}
return min;
}
}