归并排序与逆序对

Posted 2020-12-11 shengunershu

《算法导论》第二章，讨论了归并排序，其中思考题2-4提出了逆序对问题，需要注意边界条件：

# inversion 逆序对
A =[3,1,4,5,2,-3,6]
def inversion(arr):
    ‘‘‘
    冒泡版
    ‘‘‘
    invs = []
    for i in range(len(arr)):
        for j in range(i):
            if arr[j] > arr[i]:
                invs.append(str(arr[j])+‘,‘+str(arr[i]))
    return invs

allinvs = []
def inversionmerge(arr):
    ‘‘‘
    归并版
    ‘‘‘
    if len(arr)==1:
        return arr
    m = int((len(arr) +1)/2)
    arr1 = inversionmerge(arr[0:m])
    arr2 = inversionmerge(arr[m:len(arr)])
    return merge(arr1,arr2)

def merge(arr1, arr2):
    arr = []
    i1, i2 = 0, 0
    l = len(arr1) + len(arr2)
    for k in range(l):
        if i1==len(arr1) and i2 == len(arr2):
            break
        if i1 >= len(arr1):
            arr.append(arr2[i2])
            i2+=1
            continue
        if i2 >= len(arr2):
            arr.append(arr1[i1])
            # 逆序
            for n in range(len(arr1)-1, i1+1, -1):
                allinvs.append(formatinvs(arr1[n],arr2[i2-1]))
            i1+=1
            continue
        if arr1[i1] < arr2[i2]:
            arr.append(arr1[i1])
            i1+=1
        else:
            arr.append(arr2[i2])
            # 逆序
            for m in range(len(arr1)-1, i1-1, -1):
                allinvs.append(formatinvs(arr1[m],arr2[i2]))
            i2+=1
    return arr

def formatinvs(a,b):
    return str(a)+‘,‘+str(b)

def main():
    print(inversion(A))
    inversionmerge(A)
    print(allinvs)

if __name__ == "__main__":
    main()