Array Nesting
Question
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].
Answer
Answer1
这道题看到时就打算
- 保存一个marked数组用来标记已经访问过的元素
- count和maxC用来记录最大的nestArray
然后遍历一遍，每次将访问过的mark一下，代码如下
class Solution(object):
def arrayNesting(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
marked = [0] * len(nums)
count, maxC = 0, 0
for i in range(len(nums)):
ii = i
while not marked[ii]:
marked[ii] = 1
ii = nums[ii]
count += 1
maxC = max(count, maxC)
count = 0
return maxC
Answer2
不过在discussion看到别人的解法如下
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;
}
};
仔细想一下就可以发现其实没有必要保存一个marked的数组，因为不同的nestArray实际上并不会相交，遍历的过程只需要将原数组给marked就ok了
