Java
/**
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:
N is an integer within the range [1, 20,000].
The elements of A are all distinct.
Each element of array A is an integer within the range [0, N-1].
reference: https://leetcode.com/articles/array-nesting/
*/
public class Array_Nesting {
public class Solution_skip_visited {
public int arrayNesting(int[] nums) {
boolean[] visited = new boolean[nums.length];
int res = 0;
for (int i = 0; i < nums.length; i++) {
// @note: if visited, then the count here must be less than previous visit
if (!visited[i]) {
int start = nums[i], count = 0;
do {
start = nums[start];
count++;
visited[start] = true;
}
while (start != nums[i]);
res = Math.max(res, count);
}
}
return res;
}
}
public class Solution_brute_force {
public int arrayNesting(int[] nums) {
int res = 0;
for (int i = 0; i < nums.length; i++) {
int start = nums[i], count = 0;
do {
start = nums[start];
count++;
}
while (start != nums[i]); // if not the initial start number, then continue
res = Math.max(res, count);
}
return res;
}
}
}Java
class Solution {
public int arrayNesting(int[] nums) {
int length = nums.length;
boolean[] visited = new boolean[length];
int maxLength = 0;
for (int i = 0; i < length; i++) {
if (visited[i])
continue;
int curLength = 0;
int index = i;
while (!visited[index]) {
int num = nums[index];
visited[index] = true;
index = num;
curLength++;
}
maxLength = Math.max(maxLength, curLength);
}
return maxLength;
}
}