659. Split Array into Consecutive Subsequences

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最新推荐文章于 2025-05-26 01:30:00 发布

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分类专栏： leetcode greedy

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本文介绍了一种算法，用于判断是否能将已排序且可能包含重复元素的整数数组分割为若干个至少包含3个连续整数的子序列。通过实例展示了如何使用该算法，并提供了实现代码。

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You are given an integer array sorted in ascending order (may contain duplicates), you need to split them into several subsequences, where each subsequences consist of at least 3 consecutive integers. Return whether you can make such a split.

Example 1:

Input: [1,2,3,3,4,5]
Output: True
Explanation:
You can split them into two consecutive subsequences : 
1, 2, 3
3, 4, 5

Example 2:

Input: [1,2,3,3,4,4,5,5]
Output: True
Explanation:
You can split them into two consecutive subsequences : 
1, 2, 3, 4, 5
3, 4, 5

Example 3:

Input: [1,2,3,4,4,5]
Output: False

Note:

The length of the input is in range of [1, 10000]

思路：O(N)
* O(N)就是遍历几遍啦，遍历的时候考虑每个数的去向，（这是个套路？？）
* 一个数要么加到一个已经形成的队列里，要么新开辟一个队列
* （如果再加到已经形成的队列的element的个数超过3个，就一定能形成新的队列）
* （而如果少于3个，就只能加到原有队列，所以是优先加到已经形成的队列）
* 如果都不行就说明不能形成，return false
*
* 前提是数组是有序的

public class Solution {
    public boolean isPossible(int[] nums) {
    	Map<Integer, Integer> unused_freq = new HashMap<Integer, Integer>(), need_freq = new HashMap<Integer, Integer>();
        for(int i : nums)	unused_freq.put(i, unused_freq.containsKey(i)?1+unused_freq.get(i):1);
        
        for(int i : nums) {
        	if(unused_freq.get(i) == 0)	continue; 	// 这个数用完了
        	
        	if(need_freq.containsKey(i) && need_freq.get(i)>0) {
        		// 加到已经有的队列
        		need_freq.put(i, need_freq.get(i)-1);
        		need_freq.put(i+1, need_freq.containsKey(i+1)?need_freq.get(i+1)+1:1);
        		unused_freq.put(i, unused_freq.get(i)-1);
        	} else if(unused_freq.containsKey(i+1) && unused_freq.get(i+1)>0
        			&& unused_freq.containsKey(i+2) && unused_freq.get(i+2)>0) {
        		// 加到新的队列
        		need_freq.put(i+3, need_freq.containsKey(i+3)?1+need_freq.get(i+3):1);
        		unused_freq.put(i, unused_freq.get(i)-1);
        		unused_freq.put(i+1, unused_freq.get(i+1)-1);
        		unused_freq.put(i+2, unused_freq.get(i+2)-1);
        	} else {
        		// 这个数无处容身
        		return false;
        	}
        }
        
        return true;
    }
}

遍历到某个位置的时候不能拖到后面处理，要及时出击，把当前处理干净

不然留到后面是无法处理的