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36 lines
2.0 KiB
Python
36 lines
2.0 KiB
Python
'''
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Author : Mehdi ALAOUI
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This is a pure Python implementation of Dynamic Programming solution to the longest increasing subsequence of a given sequence.
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The problem is :
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Given an ARRAY, to find the longest and increasing sub ARRAY in that given ARRAY and return it.
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Example: [10, 22, 9, 33, 21, 50, 41, 60, 80] as input will return [10, 22, 33, 41, 60, 80] as output
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'''
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def longestSub(ARRAY): #This function is recursive
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ARRAY_LENGTH = len(ARRAY)
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if(ARRAY_LENGTH <= 1): #If the array contains only one element, we return it (it's the stop condition of recursion)
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return ARRAY
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#Else
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PIVOT=ARRAY[0]
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LONGEST_SUB=[] #This array will contains the longest increasing sub array
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for i in range(1,ARRAY_LENGTH): #For each element from the array (except the pivot),
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if (ARRAY[i] < PIVOT): #if the element is smaller than the pivot, it won't figure on the sub array that contains the pivot
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TEMPORARY_ARRAY = [ element for element in ARRAY[i:] if element >= ARRAY[i] ] #But it cas figure in an increasing sub array starting from this element
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TEMPORARY_ARRAY = longestSub(TEMPORARY_ARRAY) #We calculate the longest sub array that starts from this element
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if ( len(TEMPORARY_ARRAY) > len(LONGEST_SUB) ): #And we save the longest sub array that begins from an element smaller than the pivot (in LONGEST_SUB)
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LONGEST_SUB = TEMPORARY_ARRAY
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TEMPORARY_ARRAY = [ element for element in ARRAY[1:] if element >= PIVOT ] #Then we delete these elements (smaller than the pivot) from the initial array
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TEMPORARY_ARRAY = [PIVOT] + longestSub(TEMPORARY_ARRAY) #And we calculate the longest sub array containing the pivot (in TEMPORARY_ARRAY)
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if ( len(TEMPORARY_ARRAY) > len(LONGEST_SUB) ): #Then we compare the longest array between TEMPORARY_ARRAY and LONGEST_SUB
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return TEMPORARY_ARRAY
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else: #And we return the longest one
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return LONGEST_SUB
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#Some examples
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print(longestSub([4,8,7,5,1,12,2,3,9]))
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print(longestSub([9,8,7,6,5,7])) |