LeetCode[Array]: Insert Interval-优快云博客

本文链接：https://blog.youkuaiyun.com/chfe007/article/details/41812571

本文讨论了在一个已排序的不重叠区间集合中插入一个新区间并进行必要合并的问题。通过示例展示了插入过程及合并后的结果，并提供了一种实现思路与代码示例。

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Given a set of non-overlapping intervals, insert a new interval into the intervals (merge if necessary).
You may assume that the intervals were initially sorted according to their start times.
Example 1:
Given intervals [1,3],[6,9], insert and merge [2,5] in as [1,5],[6,9].
Example 2:
Given [1,2],[3,5],[6,7],[8,10],[12,16], insert and merge [4,9] in as [1,2],[3,10],[12,16].
This is because the new interval [4,9] overlaps with [3,5],[6,7],[8,10].

看到这个题目，我首先产生的一个问题就是：是在原来的区间vector上修改，还是重新建一个然后返回？这个问题虽然不会导致算法根本上的差别，但是在实现上却会带来比较大的差别。

关于这个问题，我的想法是：由于传入的参数是引用类型，所以我认为不应该对原来的vector进行修改，而应该新建一个返回。

我的解法是设置两个标志foundStart和foundEnd分别标识是否找到新区间的起始点和终止点是否找到，纯粹地根据区间关系来确定，所以实现起来也相对比较麻烦：

vector<Interval> insert(vector<Interval> &intervals, Interval newInterval) {
    vector<Interval> mergedIntervals;

    int start = 0, end = 0;
    bool foundStart = false, foundEnd = false;
    for (auto interval : intervals) {
        if (!foundStart) {
            if (interval.end < newInterval.start) mergedIntervals.push_back(interval);
            else if (interval.start > newInterval.end) {
                mergedIntervals.push_back(newInterval);
                mergedIntervals.push_back(interval);
                foundStart = true;
                foundEnd = true;
            }
            else {
                start = interval.start < newInterval.start ? interval.start : newInterval.start;
                foundStart = true;
                if (newInterval.end <= interval.end) {
                    mergedIntervals.push_back(Interval(start, interval.end));
                    foundEnd = true;
                }
                else end = newInterval.end;
            }
        }
        else if (!foundEnd) {
            if (interval.start > end) {
                mergedIntervals.push_back(Interval(start, end));
                mergedIntervals.push_back(interval);
                foundEnd = true;
            }
            else if (end < interval.end) {
                mergedIntervals.push_back(Interval(start, interval.end));
                foundEnd = true;
            }
        }
        else mergedIntervals.push_back(interval);
    }

    if (!foundStart) mergedIntervals.push_back(newInterval);
    else if (!foundEnd) mergedIntervals.push_back(Interval(start, end));

    return mergedIntervals;
}

在Discuss上有更好的解法。