Find Minimum in Rotated Sorted Array II
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Question Solution
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.
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分析：如果沒有重複的可以直接比較就可以中間項和末尾項即可區別出使用哪一個片段，此題說明是重複項存在，因此除非極端情況下可，二分取其一處理外，其他情況兩部分都要處理
例如，一種情況
3 3 3 3 3 1 3
public class Solution {
int fm(int[] nums, int start, int end){
int l=end-start+1;
if(l==1)
return nums[start];
else {
int s=start;
int e=end;
int m=(s+e)/2;
int min=0;
if(nums[m]<nums[e])
min=fm(nums, s, m);
else
{
int m1=fm(nums, s, m);
min=fm(nums, m+1, e);
if(m1<min)
min=m1;
}
return min;
}
}
public int findMin(int[] nums) {
int x=fm(nums,0,nums.length-1);
return x;
}
}