几个简单的算法
一、 二分查找
二分查找又称折半查找,优点是比较次数少,查找速度快,平均性能好;其缺点是要求待查表为有序表,且插入删除困难。因此,折半查找方法适用于不经常变动而查找频繁的有序列表。首先,假设表中元素是按升序排列,将表中间位置记录的关键字与查找关键字比较,如果两者相等,则查找成功;否则利用中间位置记录将表分成前、后两个子表,如果中间位置记录的关键字大于查找关键字,则进一步查找前一子表,否则进一步查找后一子表。重复以上过程,直到找到满足条件的记录,使查找成功,或直到子表不存在为止,此时查找不成功。
# __Author__:oliver # __DATE__:2/19/17 class Binary_Search(object): def __init__(self, data_list): self.data_list = data_list self.begin = 0 # 开始位置的索引 self.end = len(data_list) - 1 # 结束位置的索引 self.middle = 0 def find(self, num): self.middle = (self.begin + self.end) // 2 # 中间位置的索引 if num < self.data_list[self.middle]: self.end = self.middle - 1 elif num > self.data_list[self.middle]: self.begin = self.middle + 1 else: return [num, self.middle] if self.begin > self.end: return 0 return self.find(num) def main(data_list, num): binary_search = Binary_Search(data_list) result = binary_search.find(num) print('%s was found,the index for %s' % (result[0], result[1])) if result else print("Not Found.") if __name__ == '__main__': data_list = [1, 3, 23, 50, 88, 91, 107, 200, 201] while 1: num = input("Please enter the number you want to find (e)Exit:").strip() if num == "e": break main(data_list, int(num))
二、 Python自带排序方法
三、 侏儒排序法
def gnomesort(data_list): i = 0 while i < len(data_list): if i == 0 or data_list[i-1] <= data_list[i]: i += 1 else: data_list[i],data_list[i-1] = data_list[i-1],data_list[i] i -= 1 if __name__ == '__main__': data_list = [8, 10, 9, 6, 4, 16, 5, 13, 26, 18, 2, 45, 34, 23, 1, 7, 3] print(data_list) gnomesort(data_list) print(data_list)
四、 归并排序法
def mergesort(data_list): mid = len(data_list) // 2 lft,rgt = data_list[:mid],data_list[mid:] if len(lft) > 1: lft = mergesort(lft) if len(rgt) > 1: rgt = mergesort(rgt) result = [] while lft and rgt: if lft[-1] >= rgt[-1]: result.append(lft.pop()) else: result.append(rgt.pop()) result.reverse() return (lft or rgt) + result if __name__ == '__main__': data_list = [8, 10, 9, 6, 4, 16, 5, 13, 26, 18, 2, 45, 34, 23, 1, 7, 3] print(data_list) res = mergesort(data_list) print(res)
posted on 2017-02-19 14:01 oliver.lee 阅读(368) 评论(0) 编辑 收藏 举报