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C++实现求最大间距 C++实现LeetCode(164.求最大间距)

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[LeetCode] 164. Maximum Gap 求最大间距

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:

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啊,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:

https://github.com/grandyang/leetcode/issues/164 

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

https://leetcode.com/problems/maximum-gap/discuss/50643/bucket-sort-java-solution-with-explanation-on-time-and-space

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