[leetcode] 57. Insert Interval 解题报告

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最新推荐文章于 2024-05-22 13:34:14 发布

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分类专栏： leetcode array 文章标签： array leetcode

本文链接：https://blog.youkuaiyun.com/qq508618087/article/details/50658834

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本文介绍了一个高效的算法，用于解决在已排序的不重叠区间集合中插入新区间的问题，并在必要时进行区间合并。通过两个实例展示了算法的具体应用，同时提供了实现该算法的C++代码。

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题目链接：https://leetcode.com/problems/insert-interval/

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].

思路：找出左右边界即可，然后将新的区间都copy到新容器中，时间复杂度为O(n)．

代码如下：

/**
 * 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> ans;
        int i=0, left = newInterval.start, right = newInterval.end;
        for(i =0; i < intervals.size(); i++)
        {
            if(intervals[i].end < left) ans.push_back(intervals[i]);
            else if(intervals[i].start > right) break;
            else left=min(left, intervals[i].start), right=max(right, intervals[i].end);
        }
        ans.push_back(Interval(left, right));
        for(int j = i; j < intervals.size(); j++) ans.push_back(intervals[j]);
        return ans;
    }
};