659. Split Array into Consecutive Subsequences
作者:互联网
You are given an integer array nums
that is sorted in non-decreasing order.
Determine if it is possible to split nums
into one or more subsequences such that both of the following conditions are true:
- Each subsequence is a consecutive increasing sequence (i.e. each integer is exactly one more than the previous integer).
- All subsequences have a length of
3
or more.
Return true
if you can split nums
according to the above conditions, or false
otherwise.
A subsequence of an array is a new array that is formed from the original array by deleting some (can be none) of the elements without disturbing the relative positions of the remaining elements. (i.e., [1,3,5]
is a subsequence of [1,2,3,4,5]
while [1,3,2]
is not).
Example 1:
Input: nums = [1,2,3,3,4,5] Output: true Explanation: nums can be split into the following subsequences: [1,2,3,3,4,5] --> 1, 2, 3 [1,2,3,3,4,5] --> 3, 4, 5
Example 2:
Input: nums = [1,2,3,3,4,4,5,5] Output: true Explanation: nums can be split into the following subsequences: [1,2,3,3,4,4,5,5] --> 1, 2, 3, 4, 5 [1,2,3,3,4,4,5,5] --> 3, 4, 5
Example 3:
Input: nums = [1,2,3,4,4,5] Output: false Explanation: It is impossible to split nums into consecutive increasing subsequences of length 3 or more.
Constraints:
1 <= nums.length <= 104
-1000 <= nums[i] <= 1000
nums
is sorted in non-decreasing order.
class Solution { public boolean isPossible(int[] nums) { int n = nums.length; if(n < 3) return false; PriorityQueue<Interval> pq = new PriorityQueue<>((a, b) -> a.end == b.end ? a.len - b.len : a.end - b.end); for(int i : nums) { while(!pq.isEmpty() && (pq.peek().end + 1 < i)) { if(pq.poll().len < 3) return false; } if(pq.isEmpty() || i == pq.peek().end) { pq.add(new Interval(i, 1)); } else { Interval it = pq.poll(); it.len++; it.end = i; pq.add(it); } } while(!pq.isEmpty()) { if(pq.poll().len <3) return false; } return true; } class Interval { int end; int len; public Interval(int end, int len) { this.end = end; this.len = len; } } }
https://leetcode.com/problems/split-array-into-consecutive-subsequences/discuss/130452/20ms-Java-PriorityQueue-with-Explanations
标签:pq,end,nums,into,Subsequences,Split,subsequences,split 来源: https://www.cnblogs.com/wentiliangkaihua/p/16601666.html