leetcode 565. Array Nesting

原题：

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: 6Explanation: 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].

代码如下：

int arrayNesting(int* nums, int numsSize) {    int max=0;    for(int n=0;n<numsSize;n++)    {        int index=n;        int length=0;        while(nums[index]!=-1)        {            int temp = index;            index=nums[index];            nums[temp]=-1;            length++;        }        if(length>max)            max=length;    }    return max;}

有时候还是在做算法的时候，还是不习惯对原始数据进行破坏，如果追求效率的话，破坏下比较快。

但是在做项目的时候，往往不允许。

所以。。。 有人据说用了并查集。。

我来看看什么是并查集/捂脸 

我这个菜鸡。。。