算法导论学习笔记之五--同时得到最大和最小值复杂度3（n/2）

def get_min_max(l):
    if len(l) < 1:
        return None
    if len(l) == 1:
        return (l[0], l[0])
    if l[0] < l[1]:
        min = l[0]
        max = l[1]
    else:
        min = l[1]
        max = l[0]
    end = len(l)
    if end % 2 != 0:
        end = end - 1
        left1 = True
    else:
        left1 = False
        
    for i in range(2, end, 2):
        if l[i] > l[i+1]:
            a = l[i+1]
            b = l[i]
        else:
            a = l[i]
            b = l[i+1]
        if a < min:
            min = a
        if b > max:
            max = b
        
    if left1:
        last = l[len(l) - 1]
        if last < min:
            min = last
        elif last > max:
            max = last
    return (min, max)
        
    
if __name__ == "__main__":
    l = [3, 30, 24, 45, 27, 100]
    print get_min_max(l)