leetcode.352. Data Stream as Disjoint Intervals

最新推荐文章于 2021-10-09 17:16:39 发布

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本文介绍了一种处理连续非负整数数据流的方法，通过维护一系列不相交的区间来总结已接收的数据。讨论了如何在数据流中添加新元素并更新区间总结，特别是在大量区间合并后总结区间数量远小于数据流大小的情况。

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Given a data stream input of non-negative integers a₁, a₂, ..., a_n, ..., summarize the numbers seen so far as a list of disjoint intervals.

For example, suppose the integers from the data stream are 1, 3, 7, 2, 6, ..., then the summary will be:

[1, 1]
[1, 1], [3, 3]
[1, 1], [3, 3], [7, 7]
[1, 3], [7, 7]
[1, 3], [6, 7]

Follow up:
What if there are lots of merges and the number of disjoint intervals are small compared to the data stream's size?

/**
 * Definition for an interval.
 * struct Interval {
 *     int start;
 *     int end;
 *     Interval() : start(0), end(0) {}
 *     Interval(int s, int e) : start(s), end(e) {}
 * };
 */
class SummaryRanges {
public:
    /** Initialize your data structure here. */
    SummaryRanges() {
        
    }
    
     int find(int val) {
        int vlen = vec.size();
        if (vlen == 0) {
            return -1;
        }
        int begin = 0;
        int end = vlen - 1;
        int mid;
        while (begin <= end) {
            mid = begin + (end - begin) / 2;
            // Here <= vs. >
            if (vec[mid].start > val) {
                end = mid - 1;
            }
            else {
                begin = mid + 1;
            }
        }
        // Here it's important to return end
        return end;
    }

    void addNum(int val) {
         int idx = find(val);
        if (idx < 0) {
            if (vec.size() > 0 && vec[0].start == val+1) {
                vec[0].start = val;
            }
            else {
                Interval inter(val, val);
                vec.insert(vec.begin(), inter);
            }
        }
        else {
            if (vec[idx].end == val-1) {
                vec[idx].end = val;
            }
            else if (vec[idx].end < val-1) {
                Interval inter(val, val);
                vec.insert(vec.begin()+idx+1, inter);
                idx++;
            }
            // check latter
            if (vec.size() > idx+1 && vec[idx+1].start == vec[idx].end + 1) {
                vec[idx].end = vec[idx+1].end;
                vec.erase(vec.begin()+idx+1);
            }
        }

    }
    
    vector<Interval> getIntervals() {
         return vec; 
    }
private:  
   vector<Interval> vec; 

};