// 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] >= nums[end-1] 的情况, 最小值可能在 mid 左...
如下: 1classSolution {2publicint[] searchRange(int[] nums,inttarget) {3//int[] ans = new int[2];4//ans[0]=-1;ans[1]=nums.length;5int[] ans = {nums.length,-1};6if(nums==null)returnans;7search(nums,target,0,nums.length-1,ans);8if(ans[0]==nums.length) ans[0]=-1;9...
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
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 cnt = 0; // 计算总数组中有多少个数小...
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...
//https://leetcode.com/problems/split-array-largest-sum/description/ public int splitArray(int[] nums, int m) { int start = 0; int end = 0;for (int i = 0; i < nums.length; i++) { start = Math.max(start, nums[i]); // in the end of the loop this will contain the m...
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int max = mid; while (max <= end && A[mid] == A[max]) { max++; } vs.push_back(min+1); vs.push_back(max - 1); return vs; } else if (A[mid] > target) { end = mid - 1; } else { begin = mid + 1; }
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]; ...
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; ...