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.
题意理解：successive连续的
输入：未排序数组，数组元素均为非负整数，在32位整数范围内
输出：输出排序后连续元素的最大差值
要求：线性时间/空间
法一：直观做法就是排序，之后一次比较，但时间空间效率不高!!!
1 class Solution
2 {
3 public:
4 int maximumGap(vector<int> &nums)
5 {//题意理解：successive连续的
6 //输入：未排序数组，数组元素均为非负整数，在32位整数范围内
7 //输出：输出排序后连续元素的最大差值
8 //要求：线性时间/空间
9 //直观做法就是排序，之后一次比较，但时间空间效率不高!!!
10 int count = nums.size();
11 if(count <2)
12 return 0;
13 sort(nums.begin(),nums.end());
14 int maxGap = nums[1] - nums[0];
15
16 for(int i=2; i<count; i++)
17 {
18 if(maxGap < nums[i]-nums[i-1])
19 maxGap = nums[i]-nums[i-1];
20 }
21 return maxGap;
22 }
23 };
法二：
官方提示：表示没有理解！！！
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.
参考：http://www.cnblogs.com/ganganloveu/p/4162290.html