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Maximum Gap

2015.1.23 15:00

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.

Solution:

  This here is not an O(n) solution, but O(n * log(n)). Sort the array and find the maximal gap.

  As for a linear solution, I'm searching for it...

  Discussions online said it could be solved using bucket sort, but that doesn't really satisfy O(n) time scale, right? At least I wouldn't take it for O(n).

Accepted code:

复制代码
 1 // 1AC, not an O(n) solution, though, one not yet come up with...
 2 #include <algorithm>
 3 using namespace std;
 4 
 5 class Solution {
 6 public:
 7     int maximumGap(vector<int> &num) {
 8         sort(num.begin(), num.end());
 9         int n;
10         
11         n = (int)num.size();
12         if (n < 2) {
13             return 0;
14         }
15         
16         int i;
17         int ans = abs(num[0] - num[1]);
18         for (i = 2; i < n; ++i) {
19             ans = max(ans, abs(num[i - 1] - num[i]));
20         }
21         
22         return ans;
23     }
24 private:
25     int max(const int &x, const int &y) {
26         return x > y ? x : y;
27     }
28     
29     int abs(const int &x) {
30         return x >= 0 ? x : -x;
31     }
32 };
复制代码

 

 posted on   zhuli19901106  阅读(140)  评论(0编辑  收藏  举报
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