試了一些pick one功能,隨機選取了一道題,題目如下:
153. Find Minimum in Rotated Sorted Array
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.
難度: Medium 通過率:37.9%
這道題最簡單的想法就是直接順序遍歷,因爲數組原先是排序好的,只經過一次翻轉,所以找到某個元素比其前一個元素小即是最小值,時間複雜度:O(n),代碼實現如下:
class Solution {
public:
int findMin(vector<int>& nums) {
for(int i = 1; i < nums.size(); i++) {
if(nums[i] < nums[i-1]) {
return nums[i];
}
}
return nums[0];
}
};
然而這樣效率是很低的,所以考慮能否嘗試其他做法。利用類似於二分查找的方法尋找最小值,具體實現如下,時間複雜度:O(logn)
class Solution {
public:
int findMin(vector<int>& nums) {
int l = 0, r = nums.size()-1 ;
while(l < r) {
if(nums[l] < nums[r]) return nums[l];
int mid = (l + r) / 2;
if(nums[mid] > nums[r]) {
l = mid + 1;
}
else {
r = mid;
}
}
return nums[l];
}
};