leetcode(11)

leetcode(11)

565. Array Nesting

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: 6
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:

  1. N is an integer within the range [1, 20,000].
  2. The elements of A are all distinct.
  3. Each element of A is an integer within the range [0, N-1].

class Solution {
public:
int arrayNesting(vector<int>& nums) {
int size=nums.size();
int maxlong=0;

//大循环,循环整个数组
for(int i=0;i<size;i++)
{
int nowlong=0;

//小循环,记录从当前位置形成环的长度

//nums[j]用来记录是否访问过,访问过的就不访问了,因为一定比我们记录过的nowlong要小
for(int j=i;nums[j]>=0;nowlong++)
{
int temp=nums[j];

//访问过的,将其mark为-1

nums[j]=-1;
j=temp;
}
maxlong=max(maxlong,nowlong);
}
return maxlong;
}
};




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