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  • [leetcode]1054. Distant Barcodes

    出差没有状态,啃不动前面几道硬骨头,倒着做。

    Ver.1:

    时间复杂度很美,但是无法处理 1 2 2 2 5, 1 1 1 2 3这类输入:

    #1 2 2 2 5
    #2 1 2 5 2

    #2 1 1 1 3
    #1 1 1 2 3
    #1 2 1 3 1

    class Solution:
        def rearrangeBarcodes(self, barcodes: List[int]) -> List[int]:
            lengthS =len(barcodes)
            #0,1
            if lengthS == 0 or lengthS == 1:
                return barcodes
            #regular
            barcodes = sorted(barcodes)
            ret =[]
            ret.append(barcodes[lengthS // 2])
            index = 0
            while(len(ret) < lengthS):
                #
                if len(ret) < lengthS:
                    if ret[-1] != barcodes[index]:
                        ret.append(barcodes[index])
                else:
                    return ret
                #
                if len(ret) < lengthS:
                    if ret[-1] != barcodes[- (index+1)]:
                        ret.append(barcodes[- (index+1)])
                else:
                    return ret
                #check
                if len(ret)%2 == 1:
                    index = index + 1
            return ret

    Ver.2:

     其实只要去掉末尾的判断,然后不断缩短list就行了:

    Runtime: 2480 ms, faster than 100.00% of Python3 online submissions forDistant Barcodes.
    Memory Usage: 14.4 MB, less than 100.00% of Python3 online submissions for Distant Barcodes.
     

    Submission Detail

    57 / 57 test cases passed.
    Status: 

    Accepted

    Runtime: 2480 ms
    Memory Usage: 14.4 MB
    Submitted: 1 minute ago

    Accepted Solutions Runtime Distribution

    Sorry. We do not have enough accepted submissions to show distribution chart.

    Accepted Solutions Memory Distribution

    Sorry. We do not have enough accepted submissions to show distribution chart.
    class Solution:
        def rearrangeBarcodes(self, barcodes: List[int]) -> List[int]:
            lengthS =len(barcodes)
            #0,1
            if lengthS == 0 or lengthS == 1:
                return barcodes
            #regular
            barcodes = sorted(barcodes)
            ret = []
            index = 0
            curr_lengthS = lengthS
            #init
            ret.append(barcodes[lengthS // 2])
            barcodes.remove(barcodes[lengthS // 2])
            curr_lengthS = curr_lengthS -1
            while(len(ret)<lengthS):
                #head
                if len(ret) < lengthS:
                    if ret[-1] != barcodes[index]:
                        ret.append(barcodes[index])
                        barcodes.remove(barcodes[index])
                        curr_lengthS = curr_lengthS - 1
                else:
                    return ret
                #midd
                if len(ret) < lengthS:
                    if ret[-1] != barcodes[curr_lengthS // 2]:
                        ret.append(barcodes[curr_lengthS // 2])
                        barcodes.remove(barcodes[curr_lengthS // 2])
                        curr_lengthS = curr_lengthS - 1
                else:
                    return ret
            return ret
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  • 原文地址:https://www.cnblogs.com/alfredsun/p/10928294.html
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