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Description
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.
Example 1:
Input: [3,6,9,1]
Output: 3
Explanation: The sorted form of the array is [1,3,6,9], either
(3,6) or (6,9) has the maximum difference 3.
Example 2:
Input: [10]
Output: 0
Explanation: The array contains less than 2 elements, therefore return 0.
Note:
- You may assume all elements in the array are non-negative integers and fit in the 32-bit signed integer range.
- Try to solve it in linear time/space.
遇到这类问题肯定先想到的是要给数组排序，但是题目要求是要线性的时间和空间，那么只能用桶排序或者基排序。这里用桶排序 Bucket Sort 来做，首先找出数组的最大值和最小值，然后要确定每个桶的容量，即为 (最大值 - 最小值) / 个数 + 1，在确定桶的个数，即为 (最大值 - 最小值) / 桶的容量 + 1，然后需要在每个桶中找出局部最大值和最小值，而最大间距的两个数不会在同一个桶中，而是一个桶的最小值和另一个桶的最大值之间的间距，这是因为所有的数字要尽量平均分配到每个桶中，而不是都拥挤在一个桶中，这样保证了最大值和最小值一定不会在同一个桶中，具体的证明博主也不会，只是觉得这样想挺有道理的，各位看官大神们若知道如何证明请务必留言告诉博主���，参见代码如下：
class Solution {
public:
int maximumGap(vector<int>& nums) {
if (nums.size() <= 1) return 0;
int mx = INT_MIN, mn = INT_MAX, n = nums.size(), pre = 0, res = 0;
for (int num : nums) {
mx = max(mx, num);
mn = min(mn, num);
}
int size = (mx - mn) / n + 1, cnt = (mx - mn) / size + 1;
vector<int> bucket_min(cnt, INT_MAX), bucket_max(cnt, INT_MIN);
for (int num : nums) {
int idx = (num - mn) / size;
bucket_min[idx] = min(bucket_min[idx], num);
bucket_max[idx] = max(bucket_max[idx], num);
}
for (int i = 1; i < cnt; ++i) {
if (bucket_min[i] == INT_MAX || bucket_max[i] == INT_MIN) continue;
res = max(res, bucket_min[i] - bucket_max[pre]);
pre = i;
}
return res;
}
};
Github 同步地址：
参考资料：
https://leetcode.com/problems/maximum-gap
http://blog.csdn.net/u011345136/article/details/41963051
https://leetcode.com/problems/maximum-gap/discuss/50642/radix-sort-solution-in-java-with-explanation
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