标签：subject ... contains xpl return 第一个 occurs pos 重复
A zero-indexed array A of length N contains all integers from 0 to N-1. Find and return the longest length of set S, where S[i] = {A[i], A[A[i]], A[A[A[i]]], ... } subjected to the rule below.
Suppose the first element in S starts with the selection of element A[i] of index = i, the next element in S should be A[A[i]], and then A[A[A[i]]]… By that analogy, we stop adding right before a duplicate element occurs in S.
Example 1:
Input: A = [5,4,0,3,1,6,2] Output: 4 Explanation: A[0] = 5, A[1] = 4, A[2] = 0, A[3] = 3, A[4] = 1, A[5] = 6, A[6] = 2. One of the longest S[K]: S[0] = {A[0], A[5], A[6], A[2]} = {5, 6, 2, 0}
给一组长为N，元素为[0，N-1]的整数数组A，构成一个集合。
集合S[i]要求：第一个元素为A[i]，第二个为A[A[i]]...直到重复数字出现。求S的最大长度。
解决方案：创建一个visited数组，用来判断某个元素是否被访问过就行了。
class Solution { public: int arrayNesting(vector<int>& nums) { int len = nums.size(); vector<int> visited(len, 0); int res = 0; for (int i=0; i<len; ++i) { if (visited[i]==0) { visited[i] = 1; int cnt = 1; int index = nums[i]; while (visited[index]==0) { visited[index] = 1; index = nums[index]; ++cnt; } res = max(cnt, res); } } return res; } };
标签：subject ... contains xpl return 第一个 occurs pos 重复
原文地址：http://www.cnblogs.com/Zzz-y/p/8011456.html
