659. Split Array into Consecutive Subsequences

659. Split Array into Consecutive Subsequences

Question

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 [<u>1</u>,2,<u>3</u>,4,<u>5</u>] while [1,3,2] is not).

Solution

用优先级队列储存每个可组成的列表。
优先级队列根据列表的长度排列。

用哈希表记录每个列表的尾数,和其对应的优先级队列。
遍历所有数字,如果存在当前数字num-1为尾数的队列,则获取长度最小的列表,并添加当前数字num在列表中。
然后将新的优先级队列放入哈希表中。

遍历整个哈希表中的数组,如果有数组的长度小于3,则返回false,否则返回true。

Code

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class Solution {
public boolean isPossible(int[] nums) {
HashMap<Integer, PriorityQueue<List<Integer>>> map = new HashMap<>();

for(int num : nums){
PriorityQueue<List<Integer>> last = map.getOrDefault(num - 1, null);
PriorityQueue<List<Integer>> curr = map.getOrDefault(num, new PriorityQueue<List<Integer>>((a,b) -> a.size() - b.size()));

if(last == null || last.size() == 0){
List<Integer> arr = new ArrayList<>();
arr.add(num);
curr.add(arr);
map.put(num, curr);
}
else{
List<Integer> arr = last.poll();
arr.add(num);
curr.add(arr);
map.put(num, curr);
}
}

for(int last : map.keySet()){
for(List<Integer> arr : map.get(last)){
if(arr.size() < 3) return false;
}
}
return true;

}
}

Solution 2

不需要保存整个列表,只需要保存对应末尾数字的列表长度即可。

Code

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class Solution {
public boolean isPossible(int[] nums) {
HashMap<Integer, PriorityQueue<Integer>> map = new HashMap<>();

for(int num : nums){
PriorityQueue<Integer> last = map.getOrDefault(num - 1, null);
PriorityQueue<Integer> curr = map.getOrDefault(num, new PriorityQueue<Integer>());

if(last == null || last.size() == 0){
curr.add(1);
map.put(num, curr);
}
else{
int min = last.poll();
curr.add(min+1);
map.put(num, curr);
}
}

for(int last : map.keySet()){
for(int len : map.get(last)){
if(len < 3) return false;
}
}
return true;

}
}
Author

Xander

Posted on

2022-08-21

Updated on

2022-08-21

Licensed under

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