Given a data stream input of non-negative integers a1, a2, ..., an, ..., 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?

NOTE: input types have been changed on April 15, 2019. Please reset to default code definition to get new method signature.

Credits:
Special thanks to @yunhong for adding this problem and creating most of the test cases.

这道题说有个数据流每次提供一个数字，然后让我们组成一系列分离的区间，这道题跟之前那道 Insert Interval 很像，思路也很像，每进来一个新的数字 val，都生成一个新的区间 [val, val]，并且新建一个空的区间数组 res，用一个变量 cur 来保存要在现有的区间数组中加入新区间的位置。此时遍历现有的区间数组 intervals，对于每一个遍历到的当前区间 interval，假如要加入的区间的结尾位置加1比当前区间的起始位置小，说明二者不相连，将当前区间加入 res。否则当要加入区间的起始位置大于当前位置的结束位置加1，说明二者也没有交集，可以将当前区间加入 res，不过此时 cur 要自增1，因为要加入区间的位置在当前区间的后面。再否则的话，二者就会有交集，需要合并，此时用二者起始位置中较小的更新要加入区间的起始位置，同理，用二者结束位置中较大的去更新要加入区间的结束位置。最终将要加入区间放在 res 中的 cur 位置，然后将 res 赋值给 intervals 即可，参见代码如下:

解法一：

class SummaryRanges {
public:
    SummaryRanges() {}
    
    void addNum(int val) {
        vector<int> newInterval{val, val};
        vector<vector<int>> res;
        int cur = 0;
        for (auto interval : intervals) {
            if (newInterval[1] + 1 < interval[0]) {
                res.push_back(interval);
            } else if (newInterval[0] > interval[1] + 1) {
                res.push_back(interval);
                ++cur;
            } else {
                newInterval[0] = min(newInterval[0], interval[0]);
                newInterval[1] = max(newInterval[1], interval[1]);
            }
        }
        res.insert(res.begin() + cur, newInterval);
        intervals = res;
    }
    vector<vector<int>> getIntervals() {
        return intervals;
    }
private:
    vector<vector<int>> intervals;
};

感谢热心网友 greentrail 的提醒，我们可以对上面的解法进行优化。由于上面的方法每次添加区间的时候，都要把 res 赋值给 intervals，整个区间数组都要进行拷贝，十分的不高效。这里换一种方式，用一个变量 overlap 来记录所有跟要加入区间有重叠的区间的个数，用变量i表示新区间要加入的位置，这样只要最后 overlap 大于0了，现在 intervals 中将这些重合的区间删掉，然后再将新区间插入，这样就不用进行整体拷贝了，提高了效率，参见代码如下：

解法二：

class SummaryRanges {
public:
    SummaryRanges() {}
    
    void addNum(int val) {
        vector<int> newInterval{val, val};
        int i = 0, overlap = 0, n = intervals.size();
        for (; i < n; ++i) {
            if (newInterval[1] + 1 < intervals[i][0]) break; 
            if (newInterval[0] <= intervals[i][1] + 1) {
                newInterval[0] = min(newInterval[0], intervals[i][0]);
                newInterval[1] = max(newInterval[1], intervals[i][1]);
                ++overlap;
            }
        }
        if (overlap > 0) {
            intervals.erase(intervals.begin() + i - overlap, intervals.begin() + i);
        }
        intervals.insert(intervals.begin() + i - overlap, newInterval);
    }
    vector<vector<int>> getIntervals() {
        return intervals;
    }
private:
    vector<vector<int>> intervals;
};

Github 同步地址：

#352

类似题目：

Insert Interval

Range Module

Find Right Interval 

Summary Ranges

参考资料:

https://leetcode.com/problems/data-stream-as-disjoint-intervals/

https://leetcode.com/problems/data-stream-as-disjoint-intervals/discuss/82557/Very-concise-c%2B%2B-solution.

https://leetcode.com/problems/data-stream-as-disjoint-intervals/discuss/82616/C%2B%2B-solution-using-map.-O(logN)-per-adding.

https://leetcode.com/problems/data-stream-as-disjoint-intervals/discuss/82553/Java-solution-using-TreeMap-real-O(logN)-per-adding.

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