How to solve LeetCode problem 565: Array Nesting efficiently?

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描述：给定一个长度为N的从0到N-1的整数数组A，找到并返回集合S的最长长度，其中S[i]={A[i], A[A[i]], A[A[A[i]]], ...}，并遵循以下规则。假设S的第一个元素为star。

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.
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 A is an integer within the range [0, N-1].

分析

• 其实对于遍历过的数字，我们不用再将其当作开头来计算了，而是只对于未遍历过的数字当作嵌套数组的开头数字，
• 不过在进行嵌套运算的时候，并不考虑中间的数字是否已经访问过，而是只要找到和起始位置相同的数字位置，
• 然后更新结果res。

代码

class Solution {
public:
int arrayNesting(vector<int>& nums) {
int n=nums.size();
int res=0;
vector<bool> visited(n,false);
for(int i=0;i<n;i++){
if(visited[i]){
continue;
}
res=max(res,solve(nums,i,visited));
}
return res;
}

int solve(vector<int>& nums,int start, vector<bool>& visited){
int i=start;
int cnt=0;
while(cnt==0||i!=start){
visited[i]=true;
i=nums[i];
++cnt;
}
return cnt;
}
};

参考文献

​​[LeetCode] Array Nesting 数组嵌套​​