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.
Notice
You may assume all elements in the array are non-negative integers and fit in the 32-bit signed integer range.
Example
Given [1, 9, 2, 5], the sorted form of it is[1, 2, 5, 9], the maximum gap is between 5and 9 = 4.
Challenge
Sort is easy but will cost O(nlogn) time. Try to solve it in linear time and space.
LeetCode上的原题，请参见我之前的博客Maximum Gap。
class Solution { public: /** * @param nums: a vector of integers * @return: the maximum difference */ int maximumGap(vector<int> nums) { if (nums.empty()) return 0; int mx = INT_MIN, mn = INT_MAX, n = nums.size(); for (int d : nums) { mx = max(mx, d); mn = min(mn, d); } int size = (mx - mn) / n + 1; int bucket_num = (mx - mn) / size + 1; vector<int> bucket_min(bucket_num, INT_MAX); vector<int> bucket_max(bucket_num, INT_MIN); set<int> s; for (int d : nums) { 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[LintCode] Maximum Gap ，如需转载请自行联系原博主。
