[LintCode] Maximum Gap 求最大间距-白红宇

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.

Have you met this question in a real interview?

Yes

Example

Given [1, 9, 2, 5], the sorted form of it is[1, 2, 5, 9], the maximum gap is between 5and 9 = 4.

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。

class Solution {public:    /**     * @param nums: a vector of integers     * @return: the maximum difference     */    int maximumGap(vector
      
        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
       
         bucket_min(bucket_num, INT_MAX);        vector
        
          bucket_max(bucket_num, INT_MIN);        set
         
           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：，如需转载请自行联系原博主。