153. Find Minimum in Rotated Sorted Array 题目 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 min element. You may assume no duplicate exists in the array. Example ...
publicintfindMin(int[] nums) {intlen =nums.length;if(len == 1)returnnums[0];intstart = 0, end = len - 1;while(start <end) {//已经有序if(nums[start] <nums[end]) {returnnums[start]; }//int mid = (start + end) / 2;//may be overflowintmid = start + ((end - start) ...
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]; } }; 5. 6. 7. 8. 9. 10.
Find Minimum in Rotated Sorted Array I II@LeetCode Find Minimum in Rotated Sorted Array 其实直接遍历也是可以也可以AC,但是更加优化的解法应该是采用二分查找的思想。如果中间值比起点大就收缩起点,如果中间值比终点大就收缩终点,直到收缩到起点和终点相邻,也就是找到了翻转的部分。这里需要注意的是,数组可能...
153. Find Minimum in Rotated Sorted Array 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. You may assume no duplicate exists in the array. ...
Memory Usage: 37.6 MB, less than 100.00% of Java online submissions for Find Minimum in Rotated Sorted Array. class Solution { public int findMin(int[] nums) { return findMin(nums, 0, nums.length - 1); } public int findMin(int[] nums, int left, int right) { ...
Since the array is increasing first & then decreasing so the maximum element would be the last one in the increasing series & the first one in the decreasing series. SO we can ignore the increasing part and check if the next element is decreasing or not. As soon as we fin...
functionsorted_array = bucket_sort(array) % Find maximum value in the array max_value = max(array); % Initialize buckets num_buckets = max_value + 1; buckets = cell(1, num_buckets); % Distribute elements into buckets fori = 1:length(array)...
In this tutorial, Java program to find the largest number in array. Here is simple algorithm to find larget element in the array. Initialize lrg with arr[0] i.e. first element in the array. If current element is greater than lrg, then set lrg to current element. ...
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