LintCode 989: Array Nesting-优快云博客

本文介绍了一种寻找数组中特定序列的最大长度的方法。通过遍历数组并跟踪已访问元素，避免重复计算，最终找到最长的有效序列长度。

989 · Array Nesting
Algorithms
Medium
Accepted Rate
64%
Description
Solution
Notes
Discuss
Leaderboard
Record

Description
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.

Contact me on wechat to get Amazon、Google requent Interview questions . (wechat id : jiuzhang0607)

N is an integer within the range [1, 20,000].
The elements of A are all distinct.
Each element of A is an integer within the range [0, N-1].
Example
Example1

Input: [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}
Example2

Input: [0,1,2]
Output: 1

解法1：这个题目很有意思。注意每次访问如果形成了环，下次这个环上的节点都不应该访问了。

class Solution {
public:
    /**
     * @param nums: an array
     * @return: the longest length of set S
     */
    int arrayNesting(vector<int> &nums) {
        int nums_size = nums.size();
        vector<int> visited(nums_size, false);
        int max_count = 0, count = 0;
        for (int i = 0; i < nums_size; i++) {
            int index = i;
            while (!visited[index]) {
                count++;
                if (max_count < count) max_count = count;
                visited[index] = true;
                index = nums[index];
            }
            count = 0;
        }
        return max_count;
    }
};