[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

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Accepted Solutions Memory Distribution

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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