intfindMin(vector<int>&nums) {intl=0,r=nums.size()-1;if(nums.size()==0)return-1;if(nums[l]<nums[r])returnnums[l];while(l<r&&l!=r-1) {intmid=(l+r)/2;if(nums[l]<=nums[mid]) l=mid;elser=mid; }returnnums[r]; }
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] < nums[right])//例如34512,在截取子序列时候,很可能就...
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 ...
public int findMin(int[] nums) { int start = 0; int end = nums.length - 1; while (start < end) { if (nums[start] < nums[end]) { return nums[start]; } int mid = (start + end) >>> 1; //必须是大于等于,比如 nums=[9,8],mid 和 start 都指向了 9 if (nums[mid] >=...
Find the minimum element. You may assume no duplicate exists in the array. class Solution { public: int findMin(vector<int>& nums) { int left=0,right=nums.size()-1; while(left<right) { int mid=(left+right)/2; if(nums[left]<=nums[right]) ...
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
This is a follow up problem to Find Minimum in Rotated Sorted Array. Would allow duplicates affect the run-time complexity? How and why? 描述 假设按照升序排序的数组在预先未知的某个点上进行了旋转。 ( 例如,数组 [0,1,2,4,5,6,7] 可能变为 [4,5,6,7,0,1,2] )。
Insert Index in Sorted Array Write a Java program to find the index of a value in a sorted array. If the value does not find return the index where it would be if it were inserted in order. Example: [1, 2, 4, 5, 6] 5(target) -> 3(index) ...
The LARGE function can return the largest number from a list of numbers after we sorted it in descending order. Let us see how to apply this function to find the second largest value with criteria. Steps: Go to cell F7 and insert the following formula. =LARGE(IF(C4:D4=F5,C5:D10),...
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...