Find Minimum in Rotated Sorted Array II

最新推荐文章于 2022-05-26 16:47:57 发布

Nereus_Li

最新推荐文章于 2022-05-26 16:47:57 发布

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本文介绍了一种在已知数组被未知轴旋转的情况下，通过二分查找法找到数组中的最小元素的方法。考虑到数组中可能存在重复元素，算法通过查找不重复点进行优化。

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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.

解题思路:二分查找，首先由中点处查找不重复点

int Find(vector<int> &num, int first, int last)
{
	int mid = (first + last) / 2;
	int LeftIdx = mid;
	int RightIdx = mid;
	while (num[LeftIdx] == num[mid] && LeftIdx >= first)
		--LeftIdx;
	while (num[RightIdx] == num[mid] && RightIdx >= first)
		++RightIdx;
	if (LeftIdx<first){
		if (RightIdx>last)
			return num[mid];
		else
			return min(num[mid], Find(num, RightIdx, last));
	}
	else{
		if (RightIdx > last)
			return min(num[mid], Find(num, first, LeftIdx));
		else
			return min(Find(num, first, LeftIdx), min(Find(num, RightIdx, last),num[mid]));
	}
}
int findMin(vector<int> &num) {
	return Find(num, 0, num.size() - 1);
}