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565. Array Nesting 阵列嵌套

时间:2017-12-15 23:35:06      阅读:175      评论:0      收藏:0      [点我收藏+]

标签:note   集合   dup   function   span   each   xpl   ++   print   


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].


  1. /**
  2. * @param {number[]} nums
  3. * @return {number}
  4. */
  5. var arrayNesting = function (nums) {
  6.    let res = 0, set = new Set();           //因为集合S会成环,因此可以用set存储遍历过的index
  7.    for (let i = 0; i < nums.length; i++) {
  8.        let curIndex = i, count = 0;
  9.        while (!set.has(curIndex)) {
  10.            set.add(curIndex);
  11.            curIndex = nums[curIndex];
  12.            count++;
  13.        }
  14.        res = Math.max(res, count);
  15.    }
  16.    return res;
  17. };
  18. //let arr = [5, 4, 0, 3, 1, 6, 2];
  19. let arr = [1, 2, 3, 4, 5, 0]
  20. console.log(arrayNesting(arr));






565. Array Nesting 阵列嵌套

标签:note   集合   dup   function   span   each   xpl   ++   print   

原文地址:http://www.cnblogs.com/xiejunzhao/p/8045107.html

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