// 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] >=
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
[LeetCode] Find Minimum in Rotated Sorted Array II 寻找旋转有序数组的最小值之二
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...
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...
(r) // l = max(l, 1) // r = min(r, n) for i := find(l); i <= r; i = find(i + 1) { // initFa 需要开 n+1 空间(或者 n+2,如果下标从 1 开始) fa[i] = to } } //rangeFullMerged := func() bool { return find(0) == n } // 连通分量个数 ...
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]; ...
let mut max = 0; let range = (k - 1)..(nums.len() - k); for i in range { for &va in a[i].iter() { for &vb in b[nums.len() - i - 2].iter() { max = max.max(va ^ vb); } } } max } 123456789101112131415161718192021222324252627282930...
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