565. Array Nesting

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Problem statement:

A zero-indexed array A consisting of N different integers is given. The array contains all integers in the range [0, N - 1].

Sets S[K] for 0 <= K < N are defined as follows:

S[K] = { A[K], A[A[K]], A[A[A[K]]], ... }.

Sets S[K] are finite for each K and should NOT contain duplicates.

Write a function that given an array A consisting of N integers, return the size of the largest set S[K] for this array.

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}

Note:

1. N is an integer within the range [1, 20,000].
2. The elements of A are all distinct.
3. Each element of array A is an integer within the range [0, N-1].

Solution:

This is a DFS solution. I solved it by employing the DFS template with a returned length of S[k].

For each S[k], it forms a circle. It means any element in this circle returns the same length.

For the purpose of pruning, we set a visited array to denote whether current element has been visited before. 

Time complexity is O(n). 

Space complexity is O(n).

class Solution {
public:
    int arrayNesting(vector<int>& nums) {
        int largest = 0;
        vector<int> visited(nums.size(), 0);
        for(int i = 0; i < nums.size(); i++){
            largest = max(largest, largest_nesting(nums, visited, nums[i], 0));
        }
        return largest;
    }
    int largest_nesting(vector<int>& nums, vector<int>& visited, int idx, int size){
        if(visited[nums[idx]] == 0){
            visited[nums[idx]] = 1;
            return largest_nesting(nums, visited, nums[idx], size + 1);
        } else {
            return size;
        }
    }
};

565. Array Nesting

标签：element   tor   define   his   one   inpu   blog   code   font   

原文地址：http://www.cnblogs.com/wdw828/p/6917196.html