标签：
Sorting solution O(nlogn):
1 class Solution { 2 public: 3 int maximumGap(vector<int> &num) { 4 int len = num.size(), result = 0; 5 if (len < 2) return 0; 6 sort(num.begin(), num.end()); 7 for (int i = 0; i < len-1; i++){ 8 result = max(result, num[i+1] - num[i]); 9 } 10 return result; 11 } 12 };
O(n), bucket sort. The hint gets from leetcode:
Suppose there are N elements and they range from A to B.
Then the maximum gap will be no smaller than ceiling[(B - A) / (N - 1)]
Let the length of a bucket to be len = ceiling[(B - A) / (N - 1)], then we will have at most num = (B - A) / len + 1 of bucket
for any number K in the array, we can easily find out which bucket it belongs by calculating loc = (K - A) / len and therefore maintain the maximum and minimum elements in each bucket.
Since the maximum difference between elements in the same buckets will be at most len - 1, so the final answer will not be taken from two elements in the same buckets.
For each non-empty buckets p, find the next non-empty buckets q, then q.min - p.max could be the potential answer to the question. Return the maximum of all those values.
Analysis written by @porker2008.
1 class Solution { 2 public: 3 int maximumGap(vector<int> &num) { 4 int len = num.size(), result = 0, gMax = INT_MIN, gMin = INT_MAX, blen = 0, index = 0; 5 if (len < 2) return 0; 6 for (int i : num) { 7 gMin = min(gMin, i); 8 gMax = max(gMax, i); 9 } 10 blen = (gMax - gMin)/len + 1; 11 vector<vector<int> > buckets((gMax - gMin)/blen + 1); 12 for (int i : num) { 13 int range = (i - gMin)/blen; 14 if (buckets[range].empty()) { 15 buckets[range].reserve(2); 16 buckets[range].push_back(i); 17 buckets[range].push_back(i); 18 } else { 19 if (i < buckets[range][0]) buckets[range][0] = i; 20 if (i > buckets[range][1]) buckets[range][1] = i; 21 } 22 } 23 for (int i = 1; i < buckets.size(); i++) { 24 if (buckets[i].empty()) continue; 25 result = max(buckets[i][0] - buckets[index][1], result); 26 index = i; 27 } 28 return result; 29 } 30 };
LeetCode – Refresh – Maximum Gap
标签：
原文地址：http://www.cnblogs.com/shuashuashua/p/4355079.html
