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LeetCode 解题报告(387) -- 164. Maximum Gap

Problem

Given an integer array nums, return the maximum difference between two successive elements in its sorted form. If the array contains less than two elements, return 0.

You must write an algorithm that runs in linear time and uses linear extra space.

Example 1:

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Input: nums = [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:

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Input: nums = [10]
Output: 0
Explanation: The array contains less than 2 elements, therefore return 0.

Constraints:

  • 1 <= nums.length <= 104
  • 0 <= nums [i] <= 109

Analysis

   题目提示了就需要排序,但是又限定了线性的空间复杂度和时间复杂度,只能用桶排序了。要先确定桶的容量和桶的数量:

  • 桶的容量 = (最大值 - 最小值)/ 个数 + 1;
  • 桶的个数 = (最大值 - 最小值)/ 桶的容量 + 1。

   然后需要在每个桶都找到一个极大值和极小值,因为最大间距的两个数不会存在同一个桶中(这是由上面的计算方式决定的,这里没有严谨的推导),所以结果必然是后一个个桶的最小值和前一个桶的最大值的差。


Solution

   因为我们不是需要整个排序的结果,所以没有必要实现完整的桶排序,所以我们只需要记录每个桶的极大值和极小值,最后把这些极大值、极小值都处理一遍即可。


Code

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class Solution {
public:
int maximumGap(vector<int>& nums) {
if (nums.size () <= 1) {
return 0;
}
int mx = INT_MIN, mn = INT_MAX, size = nums.size ();

for (int num: nums) {
mx = max (mx, num);
mn = min (mn, num);
}

int capacity = (mx - mn) /size + 1, cnt = (mx - mn) /capacity + 1;

vector<int> bucket_min (cnt, INT_MAX), bucket_max (cnt, INT_MIN);
for (int num: nums) {
int idx = (num - mn) /capacity;
bucket_min [idx] = min (bucket_min [idx], num);
bucket_max [idx] = max (bucket_max [idx], num);
}

int pre = 0, result = 0;

for (int i = 1; i < cnt; i++) {
if (bucket_min [i] == INT_MAX || bucket_max [i] == INT_MIN) {
continue;
}

result = max (result, bucket_min [i] - bucket_max [pre]);
pre = i;
}
return result;
}
};

Summary

   这道题借用了桶排序的思路,整体来说还是比较难的,严格的证明我也还没搞懂。这道题目的分享到这里,感谢你的支持!

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