C++实现LeetCode(164.求最大间距)
[leetcode] 164. maximum gap 求最大间距
given an unsorted array, find the maximum difference between the successive elements in its sorted form.
return 0 if the array contains less than 2 elements.
example 1:
input: [3,6,9,1]
output: 3
explanation: the sorted form of the array is [1,3,6,9], either
(3,6) or (6,9) has the maximum difference 3.
example 2:
input: [10]
output: 0
explanation: the array contains less than 2 elements, therefore return 0.
note:
- you may assume all elements in the array are non-negative integers and fit in the 32-bit signed integer range.
- try to solve it in linear time/space.
遇到这类问题肯定先想到的是要给数组排序,但是题目要求是要线性的时间和空间,那么只能用桶排序或者基排序。这里用桶排序 bucket sort 来做,首先找出数组的最大值和最小值,然后要确定每个桶的容量,即为 (最大值 - 最小值) / 个数 + 1,在确定桶的个数,即为 (最大值 - 最小值) / 桶的容量 + 1,然后需要在每个桶中找出局部最大值和最小值,而最大间距的两个数不会在同一个桶中,而是一个桶的最小值和另一个桶的最大值之间的间距,这是因为所有的数字要尽量平均分配到每个桶中,而不是都拥挤在一个桶中,这样保证了最大值和最小值一定不会在同一个桶中,具体的证明博主也不会,只是觉得这样想挺有道理的,各位看官大神们若知道如何证明请务必留言告诉博主啊,参见代码如下:
class solution { public: int maximumgap(vector<int>& nums) { if (nums.size() <= 1) return 0; int mx = int_min, mn = int_max, n = nums.size(), pre = 0, res = 0; for (int num : nums) { mx = max(mx, num); mn = min(mn, num); } int size = (mx - mn) / n + 1, cnt = (mx - mn) / size + 1; vector<int> bucket_min(cnt, int_max), bucket_max(cnt, int_min); for (int num : nums) { int idx = (num - mn) / size; bucket_min[idx] = min(bucket_min[idx], num); bucket_max[idx] = max(bucket_max[idx], num); } for (int i = 1; i < cnt; ++i) { if (bucket_min[i] == int_max || bucket_max[i] == int_min) continue; res = max(res, bucket_min[i] - bucket_max[pre]); pre = i; } return res; } };
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