Given an unsorted array, find the maximum difference between the successive elements in its sorted form.
Try to solve it in linear time/space.
Return 0 if the array contains less than 2 elements.
You may assume all elements in the array are non-negative integers and fit in the 32-bit signed integer range.
Credits:
Special thanks to @porker2008 for adding this problem and creating all test cases.
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。这里我用桶排序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。代码如下：
class Solution { public: int maximumGap(vector<int> &numss) { if (numss.empty()) return 0; int mx = INT_MIN, mn = INT_MAX, n = numss.size(); for (int d : numss) { mx = max(mx, d); mn = min(mn, d); } int size = (mx - mn) / n + 1; int bucket_nums = (mx - mn) / size + 1; vector<int> bucket_min(bucket_nums, INT_MAX); vector<int> bucket_max(bucket_nums, INT_MIN); set<int> s; for (int d : numss) { int idx = (d - mn) / size; bucket_min[idx] = min(bucket_min[idx], d); bucket_max[idx] = max(bucket_max[idx], d); s.insert(idx); } int pre = 0, res = 0; for (int i = 1; i < n; ++i) { if (!s.count(i)) continue; res = max(res, bucket_min[i] - bucket_max[pre]); pre = i; } return res; } };
本文转自博客园Grandyang的博客，原文链接：求最大间距[LeetCode] Maximum Gap ，如需转载请自行联系原博主。
