154. Find Minimum in Rotated Sorted Array II
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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 an array sorted in ascending order 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.
这道题与上周那道的唯一区别在于元素可能有重复,我们仍然采用上面的方法,只是需要处理mid与start相等这种
额外情况。
nums[mid] > nums[start],右半区间查找。
nums[mid] < nums[start],左半区间查找。
nums[mid] = nums[start],出现这种情况,我们跳过start,重新查找,譬如[2,2,2,1],nums[mid] = nums[start]都为2,
这时候我们跳过start,使用[2,2,1]继续查找。
代码如下:
class Solution {
public:
int findMin(vector<int>& nums) {
int size = nums.size();
if(size == 0) {
return 0;
} else if(size == 1) {
return nums[0];
} else if(size == 2) {
return min(nums[0], nums[1]);
}
int start = 0;
int stop = size - 1;
while(start < stop - 1) {
if(nums[start] < nums[stop]) {
return nums[start];
}
int mid = start + (stop - start) / 2;
if(nums[mid] > nums[start]) {
start = mid;
}else if(nums[mid]<nums[start]){
stop =mid;
}else{
start++;
}
}
return min(nums[start], nums[stop]);
}
};
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