给你一个整数数组 nums ，和一个表示限制的整数 limit，请你返回最长连续子数组的长度，该子数组中的任意两个元素之间的绝对差必须小于或者等于 limit 。
如果不存在满足条件的子数组，则返回 0 。
示例 1：
输入：nums = [8,2,4,7], limit = 4 输出：2 解释：所有子数组如下： [8] 最大绝对差 |8-8| = 0 <= 4. [8,2] 最大绝对差 |8-2| = 6 > 4. [8,2,4] 最大绝对差 |8-2| = 6 > 4. [8,2,4,7] 最大绝对差 |8-2| = 6 > 4. [2] 最大绝对差 |2-2| = 0 <= 4. [2,4] 最大绝对差 |2-4| = 2 <= 4. [2,4,7] 最大绝对差 |2-7| = 5 > 4. [4] 最大绝对差 |4-4| = 0 <= 4. [4,7] 最大绝对差 |4-7| = 3 <= 4. [7] 最大绝对差 |7-7| = 0 <= 4. 因此，满足题意的最长子数组的长度为 2 。
示例 2：
输入：nums = [10,1,2,4,7,2], limit = 5 输出：4 解释：满足题意的最长子数组是 [2,4,7,2]，其最大绝对差 |2-7| = 5 <= 5 。
示例 3：
输入：nums = [4,2,2,2,4,4,2,2], limit = 0 输出：3
提示：
1 <= nums.length <= 10^51 <= nums[i] <= 10^90 <= limit <= 10^9
Method 1 sliding window
Seeking min/max for sliding window,O(n^2)
py
class Solution: def longestSubarray(self, nums: List[int], limit: int) -> int: curMax=curMin=nums[0] subArray=[] for num in nums: if abs(curMax-curMin)<=limit and abs(num-curMax)<=limit and abs(num-curMin)<=limit: curMax=max(curMax,num) curMin=min(curMin,num) subArray.append(num) else: subArray.append(num) subArray.pop(0) curMax=max(subArray) curMin=min(subArray) return len(subArray)
Java
class Solution { public int longestSubarray(int[] nums, int limit) { int curMax=nums[0]; int curMin=nums[0]; Queue<Integer> subArray=new LinkedList<>(); for(int num:nums){ if(Math.abs(curMax-curMin)<=limit&&Math.abs(num-curMax)<=limit&&Math.abs(num-curMin)<=limit){ curMax=Math.max(curMax,num); curMin=Math.min(curMin,num); subArray.offer(num); } else{ subArray.offer(num); subArray.poll(); curMax=Collections.max(subArray); curMin=Collections.min(subArray); } } return subArray.size(); } }
Method 2 use priority queue or deque
Time O(N)
Space O(N)
py_0
class Solution: def longestSubarray(self, nums: List[int], limit: int) -> int: res=1 l=0 maxd=deque() mind=deque() for r,num in enumerate(nums): # sliding window max/min while maxd and maxd[-1]<num:maxd.pop() while mind and mind[-1]>num:mind.pop() maxd.append(num) mind.append(num) # Pop out the front one by one while maxd and mind and maxd[0]-mind[0]>limit: if maxd[0]==nums[l]:maxd.popleft() if mind[0]==nums[l]:mind.popleft() l+=1 res=max(res,r-l+1) return res
py_1
class Solution: def longestSubarray(self, nums: List[int], limit: int) -> int: i=0 maxd=deque() mind=deque() for num in nums: while maxd and maxd[-1]<num:maxd.pop() while mind and mind[-1]>num:mind.pop() maxd.append(num) mind.append(num) if maxd[0]-mind[0]>limit: if maxd[0]==nums[i]:maxd.popleft() if mind[0]==nums[i]:mind.popleft() i+=1 return len(nums)-i
Java
class Solution { public int longestSubarray(int[] nums, int limit) { Deque<Integer> maxd = new ArrayDeque<>(); Deque<Integer> mind = new ArrayDeque<>(); int i = 0, j; for (j = 0; j < nums.length; j++) { while (!maxd.isEmpty() && maxd.peekLast() < nums[j]) maxd.pollLast(); while (!mind.isEmpty() && mind.peekLast() > nums[j]) mind.pollLast(); maxd.add(nums[j]); mind.add(nums[j]); if (maxd.peek() - mind.peek() > limit) { if (maxd.peek() == nums[i]) maxd.poll(); if (mind.peek() == nums[i]) mind.poll(); i++; } } return j - i; } }
cpp
class Solution { public: int longestSubarray(vector<int>& nums, int limit) { deque<int> maxd,mind; int i = 0, j; for (j = 0; j < nums.size(); j++) { while (!maxd.empty() && maxd.back() < nums[j]) maxd.pop_back(); while (!mind.empty() && mind.back() > nums[j]) mind.pop_back(); maxd.push_back(nums[j]); mind.push_back(nums[j]); if (maxd.front() - mind.front() > limit) { if (maxd.front() == nums[i]) maxd.pop_front(); if (mind.front() == nums[i]) mind.pop_front(); i++; } } return j - i; } };
Method 3 binary search and remove
Keep an increasing list q.
Binary insert the current element.
If
q[-1]-q[0]>limit
,
binary search the position of nums[i] and remove it from the list.
Time O(N^2)
Space O(N)
class Solution: def longestSubarray(self, nums: List[int], limit: int) -> int: i=0 q=[] for j in range(len(nums)): bisect.insort(q,nums[j]) if q[-1]-q[0]>limit: q.pop(bisect.bisect(q,nums[i])-1) i+=1 return j-i+1
Method 4 use two heaps
Time O(NogN)
Space O(N)
class Solution: def longestSubarray(self, nums: List[int], limit: int) -> int: maxq, minq = [], [] res = i = 0 for j, a in enumerate(nums): heapq.heappush(maxq, [-a, j]) heapq.heappush(minq, [a, j]) while -maxq[0][0] - minq[0][0] > limit: i = min(maxq[0][1], minq[0][1]) + 1 while maxq[0][1] < i: heapq.heappop(maxq) while minq[0][1] < i: heapq.heappop(minq) res = max(res, j - i + 1) return res
Method 5 use TreeMap
Use one tree map can easily get the maximum and the minimum at the same time.
In java, we can use TreeMap to count elements.
In cpp, it suports multi treeset, that's even better.
Time O(NogN)
Space O(N)
Java
class Solution { public int longestSubarray(int[] nums, int limit) { int i=0,j; TreeMap<Integer,Integer> m=new TreeMap<>(); for(j=0;j<nums.length;j++){ m.put(nums[j],1+m.getOrDefault(nums[j],0)); if(m.lastEntry().getKey()-m.firstEntry().getKey()>limit){ m.put(nums[i],m.get(nums[i])-1); if(m.get(nums[i])==0) m.remove(nums[i]); i++; } } return j-i; } }
cpp
class Solution { public: int longestSubarray(vector<int>& nums, int limit) { int i=0,j; multiset<int>m; for(j=0;j<nums.size();j++){ m.insert(nums[j]); if(*m.rbegin()-*m.begin()>limit) m.erase(m.find(nums[i++])); } return j-i; } };