[LeetCode]Maximum Gapd

最新推荐文章于 2024-10-15 15:32:42 发布

Three_Thousand_World

最新推荐文章于 2024-10-15 15:32:42 发布

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本文链接：https://blog.youkuaiyun.com/Three_Thousand_World/article/details/47286561

本文介绍了一种使用桶排序算法解决寻找数组中最大间隔的方法。通过将元素分布到若干个桶中，并维护每个桶的最大和最小值，最终找出最大间隔。文章详细解释了算法步骤并提供了Java代码实现。

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一题一句：

        int bucket_size = Math.max(1,(Max-Min)/(N) + 1);//ceil( (Max-Min)/(N-1) ) = (Max-Min-1)/(N-1) + 1
        or int  bucket_size = (int)Math.ceil( ((double)(Max-Min))/(N-1) );
	int bucket_count = (Max-Min)/bucket_size + 1;

下面用桶排序实现，这也是leetcode上给出的参考解法：

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.

补充：

Java array is a pain in ass especially after you have tasted python list. But still, need to get used to it.

//latest version
//刷题确实是有用的，至少到现在，对于java data structure 已经很熟练了
public int maximumGap(int[] nums) {
        if(nums.length < 2)
            return 0;
        int min = nums[0], max = nums[0];
        for(int n : nums){
            min = Math.min(min, n);
            max = Math.max(max, n);
        }
        if(min == max)
            return 0;
        int len = nums.length;
        //doesn't matter how much exactly, just make sure smaller than (int)Math.ceil((double)(max-min)/(len-1))
        //here use ceil because return value is int
        int bucketSize = (int)Math.ceil((double)(max-min)/(len-1));// or bucketSize -= 10;
        ArrayList all = new ArrayList();
        //doesn't matter how many exactly, just make sure larger than (max-min)/bucketSize
        int bucket_count = (max-min)/bucketSize + 1000000;
        for(int i=0; i> buckets = new ArrayList>();
        for(int i=0; i<= bucket_count; i++){
            List temp = new ArrayList();
            buckets.add(temp);
        }
        for(int i=0; i bucket = buckets.get(ind);
                bucket.set(0, Math.min(bucket.get(0),nums[i]));
                bucket.set(1, Math.max(bucket.get(1),nums[i]));
            }
        }
        int res = 0;
        int prev = buckets.get(0).get(1);
        for(int i=1; i bucket_used = new HashSet();
        for(int i=0;i