565. Array Nesting

Medium (Trung bình) C++ 🔗 Xem trên LeetCode

📋 Đề Bài

You are given an integer array nums of length n where nums is a permutation of the numbers in the range [0, n - 1].

You should build a set s[k] = {nums[k], nums[nums[k]], nums[nums[nums[k]]], ... } subjected to the following rule:

  • The first element in s[k] starts with the selection of the element nums[k] of index = k.
  • The next element in s[k] should be nums[nums[k]], and then nums[nums[nums[k]]], and so on.
  • We stop adding right before a duplicate element occurs in s[k].

Return the longest length of a set s[k].

 

Example 1:

Input: nums = [5,4,0,3,1,6,2]
Output: 4
Explanation: 
nums[0] = 5, nums[1] = 4, nums[2] = 0, nums[3] = 3, nums[4] = 1, nums[5] = 6, nums[6] = 2.
One of the longest sets s[k]:
s[0] = {nums[0], nums[5], nums[6], nums[2]} = {5, 6, 2, 0}

Example 2:

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

 

Constraints:

  • 1 <= nums.length <= 105
  • 0 <= nums[i] < nums.length
  • All the values of nums are unique.

💻 Lời Giải

C++ 0565-array-nesting.cpp 
class Solution {
public:
    int arrayNesting(vector<int>& nums) {
        int res = 0;
        int n = nums.size();
        
        for (int i = 0; i < n; ++i) {
            if (nums[i] == -1) {
                continue;
            }
            int cnt = 1, x = nums[i];
            nums[i] = -1;
            while (nums[x] != -1) {
                int temp = nums[x];
                nums[x] = -1;
                x = temp;
                cnt += 1;
            }
            res = max(res, cnt);
        }
        
        return res;
    }
};