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.
You may assume no duplicate exists in the array.
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思路:
1.对子数组sums[lo, hi],如果nums[lo] < nums[hi],则数组没有逆置,直接输出nums[lo];
2.如果nums[lo] >= nums[hi],则数组有逆置,取nums[mid]与nums[lo]比较:
1)如果nums[lo] <= nums[mid],则最少值在后半部分,这时令 lo = mid + 1,
2)否则,令 hi = mid;
3.循环1、2直接lo >= hi。
code:
class Solution {
public:
int findMin(vector<int>& nums) {
int lo = 0, hi = nums.size() - 1;
while(lo < hi) {
if(nums[lo] < nums[hi]) return nums[lo];
int mid = (lo + hi) >> 1;
if(nums[lo] <= nums[mid])
lo = mid + 1;
else
hi = mid;
}
return nums[lo];
}
};