Array Nesting

最新推荐文章于 2019-03-07 20:58:41 发布

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最新推荐文章于 2019-03-07 20:58:41 发布

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分类专栏： leetcode java

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本文介绍了一种算法，用于找出由特定规则定义的最大无重复元素序列。通过遍历数组并跟踪已访问元素，该方法能够有效地确定给定数组中最大的合法序列长度。

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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].

题目大体意思解析：用索引所在的数值作为新的下标输入，比如Ａ［０］＝５，把５作为新的数组下标输入，得到６，依次类推最后形成一个回环｛５,６,２,０｝

关键所在：将访问过的下标所在值换成－１；因为题目中说所有值都不同，所以每条路线都是唯一的，下次如果还能访问到，只能说明不是最长的，所以也没什么价值了，标成－１，没什么问题

 int max=0;
	   
	   for(int i=0;i<nums.length;i++){
		   int count=0;
		   int j=i;
		   while(nums[j]!=i&&nums[j]>0){
			   count++;
			   int x=j;
			   j=nums[j];
			   nums[x]=-1;
		   }
		   if(count>max){
			   max=count;
		   }
	   }
	   
	   return max+1;

A[0] = 5