697. Degree of an Array 数组的度

Given a non-empty array of non-negative integers nums, the degree of this array is defined as the maximum frequency of any one of its elements.

Your task is to find the smallest possible length of a (contiguous) subarray of nums, that has the same degree as nums.

Example 1:

Input: [1, 2, 2, 3, 1]
Output: 2
Explanation: 
The input array has a degree of 2 because both elements 1 and 2 appear twice.
Of the subarrays that have the same degree:
[1, 2, 2, 3, 1], [1, 2, 2, 3], [2, 2, 3, 1], [1, 2, 2], [2, 2, 3], [2, 2]
The shortest length is 2. So return 2.

Example 2:

Input: [1,2,2,3,1,4,2]
Output: 6

Note:

  • nums.length will be between 1 and 50,000.
  • nums[i] will be an integer between 0 and 49,999.

    • 给定一个非空数组的非负整数nums，该数组的“度”被定义为其任何一个元素的最大频率。 
    你的任务是找到num的子数组的最小可能长度，子数组与nums的“度”相同。
    1. /**
    2.  * @param {number[]} nums
    3.  * @return {number}
    4.  */
    5. var findShortestSubArray = function(nums) {
    6.     if (nums.length == 0 || !nums) {
    7.         return 0;
    8.     }
    9.     let m = {};
    10.     let maxTimes = 0;
    11.     for (let i = 0; i < nums.length; i++) {
    12.         let num = nums[i];
    13.         let item = m[num];
    14.         if (item) {
    15.             item.right = i;
    16.             item.times++;
    17.             maxTimes = Math.max(item.times, maxTimes);
    18.         } else {
    19.             m[num] = {
    20.                 times: 1,
    21.                 left: i,
    22.                 right: i
    23.             }
    24.             maxTimes = Math.max(1, maxTimes);
    25.         }
    26.     }
    27.     let min = 2147483647;
    28.     for (let i in m) {
    29.         let item = m[i];
    30.         if (item.times == maxTimes) {
    31.             if (item.left == item.right) {
    32.                 min = Math.min(1, min);
    33.             } else {
    34.                 min = Math.min(item.right - item.left + 1, min);
    35.             }
    36.         }
    37.     }
    38.     return min;
    39. };





    posted @ 2017-10-15 21:55  xiejunzhao  阅读(175)  评论(0)  编辑  收藏  举报