今天在网上看到一个python脚本,用来求一串数字中重复得那个数字,具体看这里:http://www.keithschwarz.com/interesting/code/?dir=find-duplicate
他得程序是
def findArrayDuplicate(array): assert len(array) > 0 # The "tortoise and hare" step. We start at the end of the array and try # to find an intersection point in the cycle. slow = len(array) - 1 fast = len(array) - 1 # Keep advancing 'slow' by one step and 'fast' by two steps until they # meet inside the loop. while True: slow = array[slow] fast = array[array[fast]] # print 'step: ', 's :',slow, ' f: ', fast if slow == fast: # print 'slow: ',slow break # Start up another pointer from the end of the array and march it forward # until it hits the pointer inside the array. finder = len(array) - 1 while True: slow = array[slow] finder = array[finder] # If the two hit, the intersection index is the duplicate element. if slow == finder: return slow
思路就是根据f(i)=A(i)这个函数,将问题抽象成一个闭环求交点得问题。不过我有点疑问,第一个while还好理解,第二while在我看来好像有死循环得可能。另外,只要是有限个数得数组,根据其抽象都会形成一个环,比如[0,1,3,2], ...f(3)=2, f(2)=3...., so, 这个程序对错误数据毫不容错。
ps。上面是我想错了,第二个while那么做是有理论依据的, 如下:
关于找到找到环的入口点的问题,当fast若与slow相遇时,slow肯定没有走遍历完链表,而fast已经在环内循环了n圈(1<=n)。假设slow走了s步,则fast走了2s步(fast步数还等于s 加上在环上多转的n圈),设环长为r,则:
2s = s + nr
则: s = nr
设整个链表长L,入口环与相遇点距离为x,起点到环入口点的距离为a。
a + x = nr
a + x = (n-1) r + r = (n-1)r + L - a
a = (n-1)r + (L - a - x)
(L – a – x)为相遇点到环入口点的距离,由此可知,从链表头到环入口点等于(n-1)循环内环+相遇点到环入口点,于是我们从链表头、与相遇点分别设一个指针,每次各走一步,两个指针必定相遇,且相遇第一点为环入口点。