[LeetCode] 565. Array Nesting 数组嵌套
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}
解法 ：
这个题实际上是找到最大的环。要注意的一点是，一旦在前一个不同的环中被访问过，它就不会出现在当前环中，可以忽略。
The idea is to, start from every number, find circles in those index-pointer-chains, every time you find a set (a circle) mark every number as visited (-1) so that next time you won't step on it again.
Java:
public class Solution {
public int arrayNesting(int[] a) {
int maxsize = 0;
for (int i = 0; i < a.length; i++) {
int size = 0;
for (int k = i; a[k] >= 0; size++) {
int ak = a[k];
a[k] = -1; // mark a[k] as visited;
k = ak;
}
maxsize = Integer.max(maxsize, size);
}
return maxsize;
}
}
Python:
class Solution(object):
def arrayNesting(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
ans, step, n = 0, 0, len(nums)
seen = [False] * n
for i in range(n):
while not seen[i]:
seen[i] = True
i, step = nums[i], step + 1
ans = max(ans, step)
step = 0
return ans
C++:
class Solution {
public:
int arrayNesting(vector<int>& a) {
size_t maxsize = 0;
for (int i = 0; i < a.size(); i++) {
size_t size = 0;
for (int k = i; a[k] >= 0; size++) {
int ak = a[k];
a[k] = -1; // mark a[k] as visited;
k = ak;
}
maxsize = max(maxsize, size);
}
return maxsize;
}
};
