leetcode 164: Maximum Gap

最新推荐文章于 2021-02-26 15:12:35 发布

mikefan1991

最新推荐文章于 2021-02-26 15:12:35 发布

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本文介绍了一种用于寻找整数数组中最大数值间隔的高效算法。该算法通过将数据分布到多个桶(bucket)中来实现，每个桶记录其内部的最大值和最小值。这种方法避免了直接排序，减少了计算复杂度。

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Very hard problem, if I didn't learn the algorithm first, I will never think of it. The idea is to divide the range max-min into some buckets. And assign all numbers to those buckets, then those numbers are already in order in the buckets. Finally scan the buckets and find out the maximum gap.

class Solution {
public:
    int maximumGap(vector<int>& nums) {
        if(nums.size()<2)
            return 0;
        int len=nums.size();
        int Max=INT_MIN,Min=INT_MAX;
        for(int i=0;i<len;i++)
        {
            if(nums[i]>Max)
                Max=nums[i];
            if(nums[i]<Min)
                Min=nums[i];
        }
        int bsize=ceil((double)(Max-Min)/(len-1));//bucket size
        int bnum=ceil((double)(Max-Min)/bsize)+1;//number of buckets
        vector<pair<int,int> > buckets(bnum,make_pair(INT_MIN,INT_MAX));
        for(int i=0;i<len;i++)
        {
            int idx=(nums[i]-Min)/bsize;//get the index of bucket this number should be
            if(buckets[idx].first<nums[i])//update the max of the bucket
                buckets[idx].first=nums[i];
            if(buckets[idx].second>nums[i])//update the min of the bucket
                buckets[idx].second=nums[i];
        }
        int res=INT_MIN,pre=0;
        for(int curr=1;curr<buckets.size();curr++)
        {
            if(buckets[curr].first==INT_MIN)//empty bucket
                continue;
            res=max(res,buckets[curr].second-buckets[pre].first);
            pre=curr;
        }
        return res;
    }
};