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。
/*** @param {number[]} nums* @return {number}*/var findShortestSubArray = function(nums) {if (nums.length == 0 || !nums) {return 0;}let m = {};let maxTimes = 0;for (let i = 0; i < nums.length; i++) {let num = nums[i];let item = m[num];if (item) {item.right = i;item.times++;maxTimes = Math.max(item.times, maxTimes);} else {m[num] = {times: 1,left: i,right: i}maxTimes = Math.max(1, maxTimes);}}let min = 2147483647;for (let i in m) {let item = m[i];if (item.times == maxTimes) {if (item.left == item.right) {min = Math.min(1, min);} else {min = Math.min(item.right - item.left + 1, min);}}}return min;};