565. Array Nesting

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;
    }
};

 

转载于:https://www.cnblogs.com/Zzz-y/p/8011456.html