【LeetCode】659. Split Array into Consecutive Subsequences 解题报告（Python）

标签（空格分隔）： LeetCode

作者：负雪明烛
id： fuxuemingzhu
个人博客： http://fuxuemingzhu.cn/

题目地址：https://leetcode.com/problems/split-array-into-consecutive-subsequences/description/

题目描述：

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]

题目大意

把一个升序的数组，分割成几个连续的递增的整数序列。如果能分割，且分割后的每个序列的长度都至少为3，那么认为成功，否则失败。

解题方法

这就是所谓的扑克牌算法，必须全部弄成“顺子”。一个“顺子”至少3张连续的牌。方法是使用优先级队列，优先把当前的牌放入到更短的“顺子”里（贪心）。

这个题的思想就是贪心+优先级队列

首先判断以（num-1）为结尾的序列是否存在，

如果存在，获取长度最小值len并出栈，创建以num为结尾的数组，并设置长度为len + 1，推入优先队列；

如果不存在，创建新的序列，以num为结尾，并且长度为1，推入优先队列，创建新的键值对（num，currentPriorityQueue）推入map中。

1，2，3，3，4，4，5，5

num last            len     current         map

1   null->(0,[ ])   0       (1, [1])    (0,[ ] ) (1, [1])

2   (1, [1])        1       (2, [2])    (0,[ ] ) (1, [ ])(2, [2])

3   (2, [2])        2       (3, [3])    (0,[ ] ) (1, [ ])(2, [ ])(3, [3])

3   (2, [ ])        0       (3, [1])    (0,[ ] ) (1, [ ])(2, [ ])(3, [3])(3, [1])

4   (3, [1])        1       (4, [2])    (0,[ ] ) (1, [ ])(2, [ ])(3, [3])(3, [ ])(4, [2])

4   (3, [3])        3       (4, [4])    (0,[ ] ) (1, [ ])(2, [ ])(3, [ ])(3, [ ])(4, [2])(4, [4])

5   (4, [2])        2       (5, [3])    (0,[ ] ) (1, [ ])(2, [ ])(3, [ ])(3, [ ])(4, [ ])(4, [4])(5, [3])

5   (4, [4])        4       (5, [5])    (0,[ ] ) (1, [ ])(2, [ ])(3, [ ])(3, [ ])(4, [ ])(4, [ ])(5, [3])(5, [5])

代码如下：

class Solution(object):

    def isPossible(self, nums):

        """

        :type nums: List[int]

        :rtype: bool

        """

        saved = collections.defaultdict(list)

        for num in nums:

            last = saved[num - 1]

            _len = 0 if (not last) else heapq.heappop(last)

            current = saved[num]

            heapq.heappush(current, _len + 1)

        for values in saved.values():

            for v in values:

                if v < 3:

                    return False

        return True

参考资料：

https://blog.csdn.net/sunday0904/article/details/78174122

日期

2018 年 8 月 29 日 ———— 还是要早起才行啊！

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【LeetCode】659. Split Array into Consecutive Subsequences 解题报告（Python）

【LeetCode】659. Split Array into Consecutive Subsequences 解题报告（Python）

题目描述：

题目大意

解题方法

日期

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