Find Minimum in Rotated Sorted Array 2 寻找旋转有序数组的最小值之二
2014-10-21 15:45
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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 a sorted array is rotated at some pivot unknown to you beforehand.
(i.e.,
Find the minimum element.
The array may contain duplicates.
这道寻找旋转有序重复数组的最小值是对之前那道问题的延伸,当数组中存在大量的重复数字时,就会破坏二分查找法的机制,我们无法取得O(lgn)的时间复杂度,又将会回到简单粗暴的O(n),比如如下两种情况:
{2, 2, 2, 2, 2, 2, 2, 2, 0, 1, 1, 2} 和 {2, 2, 2, 0, 2, 2, 2, 2, 2, 2, 2, 2}, 我们发现,当第一个数字和最后一个数字,还有中间那个数字全部相等的时候,二分查找法就崩溃了,因为它无法判断到底该去左半边还是右半边。这种情况下,我们只能回到最原始的那种简单粗暴的方法,把整个数组从头到尾撸一遍,找出最小的那个。代码如下:
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 7might become
4 5 6 7 0 1 2).
Find the minimum element.
The array may contain duplicates.
这道寻找旋转有序重复数组的最小值是对之前那道问题的延伸,当数组中存在大量的重复数字时,就会破坏二分查找法的机制,我们无法取得O(lgn)的时间复杂度,又将会回到简单粗暴的O(n),比如如下两种情况:
{2, 2, 2, 2, 2, 2, 2, 2, 0, 1, 1, 2} 和 {2, 2, 2, 0, 2, 2, 2, 2, 2, 2, 2, 2}, 我们发现,当第一个数字和最后一个数字,还有中间那个数字全部相等的时候,二分查找法就崩溃了,因为它无法判断到底该去左半边还是右半边。这种情况下,我们只能回到最原始的那种简单粗暴的方法,把整个数组从头到尾撸一遍,找出最小的那个。代码如下:
class Solution { public: int findMin(vector<int> &num) { if (num.size() <= 0) return 0; if (num.size() == 1) return num[0]; int left = 0, right = num.size() - 1; if (num[left] > num[right]) { while (left != (right - 1)) { int mid = (left + right) / 2; if (num[left] <= num[mid]) left = mid; else right = mid; } return min(num[left], num[right]); } else if (num[left] == num[right]) { int res = num[0]; for (int i = 1; i < num.size(); ++i) { res = min(res, num[i]); } return res; } return num[0]; } };
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