154. Find Minimum in Rotated Sorted Array II Hard
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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.
思路:本题思路主要是使用分治思想,想要找整个数列的最小值,就是找到小于每一个数的值,因此我们把原来的数组划分为小于某一个数pivot的小数组,每个小数组用同样的想法进行划分,直到小数组大小为0,这样用于划分的数pivot就是我们需要的数。pivot的取法类似于快排。
int min(vector<int> &nums, int pivot) { vector<int>left; for (int i = 0; i < nums.size(); i++) { if (nums[i] < pivot) { left.push_back(nums[i]); } } if (left.size() == 0) { return pivot; } return min(left, left[0]);}class Solution {public: int findMin(vector<int>& nums) { if (nums.size() == 1) { return nums[0]; } return min(nums, nums[0]); }};
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