Find Minimum in Rotated Sorted Array II(找到有序数组旋转后的最小值 II)
【难度:Medium】
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题的扩展,在原题基础上允许数组中出现重复数字。
解题思路
使用二分查找时,在153题的基础上增加中间值和末端值相等的情况时的处理。同样使用常规查找时不需改变。
c++代码如下:
//二分查找
class Solution {
public:
int findMin(vector<int>& nums) {
int low = 0;
int high = nums.size()-1;
while (low < high) {
int mid = (low+high)/2;
if (nums[mid] > nums[high])
low = mid+1;
else if (nums[mid] < nums[high])
high = mid;
else
high--;
}
return nums[high];
}
};
//常规查找
class Solution {
public:
int findMin(vector<int>& nums) {
int min = INT_MAX;
for (int i = 0; i < nums.size(); i++)
if (min > nums[i])
min = nums[i];
return min;
}
};