LeetCode-Insert Interval

最新推荐文章于 2025-08-21 20:43:22 发布

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最新推荐文章于 2025-08-21 20:43:22 发布

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分类专栏： Cracking LeetCode Interval 文章标签： leetcode Insert Interval

本文链接：https://blog.youkuaiyun.com/HolyDominat/article/details/38542767

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本文介绍了一种在已排序的非重叠区间集中插入新区间并进行必要合并的方法。通过两个示例说明了如何处理区间重叠的情况，并给出了一段C++实现代码。

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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].

Solution:

Code:

<span style="font-size:14px;">/**
 * 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 Solution {
public:
    vector<Interval> insert(vector<Interval> &intervals, Interval newInterval) {
        vector<Interval> result;
        const int length = intervals.size();
        if (length == 0) {
            result.push_back(newInterval);
            return result;
        }
        if (newInterval.end < intervals[0].start) {
            intervals.insert(intervals.begin(), newInterval);
            return intervals;
        }
        if (newInterval.start > intervals[length-1].end) {
            intervals.push_back(newInterval);
            return intervals;
        }
        int index = 0;
        for (; index < length; ++index)
            if (intervals[index].end >= newInterval.start) break;
            else result.push_back(intervals[index]);
        newInterval.start = min(intervals[index].start, newInterval.start);
        for (; index < length; ++index)
            if (intervals[index].start > newInterval.end) break;
        newInterval.end = max(intervals[index-1].end, newInterval.end);
        result.push_back(newInterval);
        if (index != length)
            result.insert(result.end(), intervals.begin()+index, intervals.end());
        return result;
    }
};</span>