565. Array Nesting

这道题目的大概意思是，先选定数组中一个起始的位置，再根据她的值定位到相应的下标，继续下去，直到出现循环为止，最后找出最长的不循环的。

显然需要将数组中每个位置起始的都要计算一遍，所以首先想到的就是dfs。

class Solution(object):    def arrayNesting(self, nums):        """        :type nums: List[int]        :rtype: int        """        if not nums:            return 0        visited = {}        result = 0        for idx in range(0, len(nums)):            if idx not in visited:                result = max(result, self.helper(nums, idx, visited))        return result    def helper(self, nums, start, visited):        """        递归移动到下一个位置，直到和起始位置一样的时候        同时记录移动的个数        :param nums:        :param start:        :param visited:        :return:        """        i = start        cnt = 0        while i != start or cnt == 0:            visited[i] = True            i = nums[i]            cnt += 1        return cnt

或者是不用函数递归的形式，直接写在一个函数里面，这样还避免了函数递归造成的消耗；

class Solution(object):    def arrayNesting(self, nums):        """        :type nums: List[int]        :rtype: int        """        visited = [False] * len(nums)        max_cnt = 0        for i in range(len(nums)):            cnt = 1            first = i            if not visited[i]:                next_idx = nums[i]  # 下一个位置                # dfs                while first != next_idx:                    visited[next_idx] = True                    next_idx = nums[next_idx]                    cnt += 1            max_cnt = max(max_cnt, cnt)        return max_cnt

转载于:https://www.cnblogs.com/George1994/p/8976748.html

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