1、题目
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.
2、分析
题意:给定一个未排序的数组。返回其排序后的数组中 相邻元素之差 最大的值。
比方给定:[5,9,8,3,15]
排序后为:[3,5,8,9,15]。相邻元素之差最大的是15-9=6,返回6。
复杂度要求:时间空间均为O(n)。
这道题最直接的解法是,先排序,得到有序数组。然后再对相邻元素作差。找出差最大的,比方以下简短的代码:
class Solution { public: int maximumGap(vector<int> &num) { if(num.size()<2) return 0; sort(num.begin(),num.end()); //O(nlogn) int gap=-1; for(int i=1;i<num.size();i++){ gap=max(gap,num[i]-num[i-1]); } return gap; } };
在Leetcode上上面的代码能够AC,但其实并没有满足时间复杂度要求。由于STL函数sort()的复杂度是O(nlogn),【sort C++ reference】。
那么。线性的排序算法有哪些?计数排序、基数排序、桶排序。
以下用桶排序实现。这也是leetcode上给出的參考解法。我直接copy过来:
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.
依据上面的思路。得到代码例如以下:class Solution { public: int maximumGap(vector<int> &num) { if (num.size() < 2) return 0; //遍历一遍。找出最大最小值 int maxNum = num[0]; int minNum = num[0]; for (int i : num) { maxNum=max(maxNum,i); minNum=min(minNum,i); } // 每一个桶的长度len,向上取整所以加+ int len = (maxNum - minNum) / num.size() + 1; //桶的个数:(maxNum - minNum) / len + 1,每一个桶里面存储属于该桶的最大值和最小值就可以,注意这里的最大最小值是局部的 vector<vector<int>> buckets((maxNum - minNum) / len + 1); for (int x : num) { int i = (x - minNum) / len; if (buckets[i].empty()) { buckets[i].reserve(2); buckets[i].push_back(x); buckets[i].push_back(x); } else { if (x < buckets[i][0]) buckets[i][0] = x; if (x > buckets[i][1]) buckets[i][1] = x; } } //gap的计算,For each non-empty buckets p, find the next non-empty buckets q, return min( q.min - p.max ) int gap = 0; int prev = 0; for (int i = 1; i < buckets.size(); i++) { if (buckets[i].empty()) continue; gap = max(gap, buckets[i][0] - buckets[prev][1]); prev = i; } return gap; } };
【其它解法以后再更新】