// 返回 nums 区间 [beg, emd) 的最小值 int findMin(const vector<int>& nums, int beg, int end) { if(beg >= end) return INT_MAX; if(beg + 1 == end) return nums[beg]; int mid = (beg+end)/2; // nums[mid] == nums[beg] >=
int findMin(vector<int>& nums) { int low=0; int high=nums.size()-1; int res=INT_MAX; while(low<high-1){ int mid=(low+high)/2; if(nums[low]<nums[mid]){ res=min(nums[low],res); low=mid+1; }else if(nums[low]>nums[mid]){ res=min(res,nums[mid]); high=mid; }else...
First, let's see the concept of 'MEDIAN' in a slightly unconventional way. That is: "if we cut the sorted array to two halves of EQUAL LENGTHS, then median is the AVERAGE OF Max(lower_half) and Min(upper_half), i.e. the
Find the minimum element. The array may contain duplicates. // [小][大] ->(rotate)-> [大][小],只有在pivot处才会有逆序 public class Solution { public int findMin(int[] nums) { int result = Integer.MAX_VALUE; int left = 0; int right = nums.length - 1; while(left <= right){...
Leetcode 1304.和为零的N个唯一整数 1 题目描述(Leetcode题目链接) 给你一个整数 n,请你返回 任意 一个由 n 个 各不相同 的整数组成的数组,并且这 n 个数相加和为 0 。 提示:1 <= n <= 1000 2 题解 此题过于开放。...Leet...
因为mid是下标,所以判断式应为cnt > mid,最后返回min 代码 public class Solution { public int findDuplicate(int[] nums) { int min = 0, max = nums.length - 1; while(min <= max){ // 找到中间那个数 int mid = min + (max - min) / 2; ...
int min = intervals[0][0], max = intervals[0][1]; for(int i = 1 ; i<intervals.length ; i++) { min = Math.min(min, intervals[i][0]); max = Math.max(max, intervals[i][1]); } int[] buckets = new int[max - min + 1]; ...
(10,Collections.reverseOrder());}publicvoidaddNum(int num){max.offer(num);min.offer(max.poll());if(max.size()<min.size())max.offer(min.poll());}publicdoublefindMedian(){if(max.size()==min.size())return(max.peek()+min.peek())/2.0;elsereturnmax.peek();}}/*class MedianFinder ...
2019-11-04 13:31 −Suppose an array of length n sorted in ascending order is rotated between 1 and n times. For example, the array... CNoodle 0 444 SET ANSI_WARNINGS 2019-12-04 16:13 −SET ANSI_WARNINGS 可以影响下列情况:设置为 ON 时,如果聚合函数(如 SUM、AVG、MAX、MIN、STDEV...
3 ii.If the second max element in the current search space is located in the first half of the array (i.e.mid > smax_all), then find the second maximum element in the first half (i.e.smax_1st = object.query(left, mid)). If the second max elements in the ent...