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. };

来自为知笔记(Wiz)

转载于:https://www.cnblogs.com/xiejunzhao/p/7674473.html