所以就不用对这种顺序情况单独讨论了。 1publicintfindMin(int[] nums)2{3returnfindMin(nums, 0, nums.length - 1);4}5//递归6publicintfindMin(int[] nums,intleft,intright)7{8if(left ==right)9{10returnnums[left];11}12if(nums[left] <
public int findMin(int[] nums) { int start = 0; int end = nums.length - 1; while (start < end) { int mid = (start + end) >>> 1; if (nums[mid] > nums[end]) { start = mid + 1; } else { end = mid; } } return nums[start]; } 解法二 解法一中我们把 mid 和end ...
1. 无重复 Suppose a sorted array is rotated at some pivot unknown to you beforehand. (i.e.,0 1 2 4 5 6 7might become4 5 6 7 0 1 2). Find the minimum element. You may assume no duplicate exists in the array. 思路:这个是找最小,和找指定数字其实差不多的。画个示意图吧 二分...
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 ...
Can you solve this real interview question? Find Minimum in Rotated Sorted Array - Suppose an array of length n sorted in ascending order is rotated between 1 and n times. For example, the array nums = [0,1,2,4,5,6,7] might become: * [4,5,6,7,0,1,2] if
Suppose a sorted array is rotated at some pivot unknown to you beforehand. 0 1 2 4 5 6 7might become4 5 6 7 0 1 2). Find the minimum element. You may assume no duplicate exists in the array. 思路: 有序数组旋转后,如果mid元素比low大,则左边有序,右边乱序,考虑最左元素是否最小元素...
Explanation: The original array was [0,1,2,4,5,6,7] and it was rotated 4 times. Java代码 publicintfindMin(int[]nums){if(nums.length==1){returnnums[0];}intleft=0;intright=nums.length-1;while(left<right){int middle=left+(right-left)/2;// 不能选和left比,因为如果整个数组升序就...
Given the sorted rotated array nums that may containduplicates, returnthe minimum element of this array. You must decrease the overall operation steps as much as possible. class Solution { public int findMin(int[] nums) { int l = 0, r = nums.length - 1; ...
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. 数列基本有序,则使用二分查找。假设数列是n,s是起始下标,e是最后下标,m是中...
Here, we are going to find the solution to find the minimum in rotated sorted array with C++ implementation.