LeetCode Algorithm 0056 - Merge Intervals (Medium)

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分类专栏： LeetCode 文章标签： LeetCode sort array 算法基础

本文链接：https://blog.youkuaiyun.com/darkrabbit/article/details/89606559

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本文详细解析了LeetCode上的经典题目——区间合并算法(Medium难度)。通过实例展示如何将所有重叠的区间进行合并，提供了一个C++实现的解决方案，并附带代码。该算法首先对输入的区间按起始点排序，然后遍历每个区间，如果当前区间的起始点大于等于前一区间的结束点，则将其添加到结果集中；反之则更新前一区间的结束点。

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LeetCode Algorithm 0056 - Merge Intervals (Medium)

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Problem Link: https://leetcode.com/problems/merge-intervals/

Description

Given a collection of intervals, merge all overlapping intervals.

Example 1:

Input: [[1,3],[2,6],[8,10],[15,18]]
Output: [[1,6],[8,10],[15,18]]
Explanation: Since intervals [1,3] and [2,6] overlaps, merge them into [1,6].

Example 2:

Input: [[1,4],[4,5]]
Output: [[1,5]]
Explanation: Intervals [1,4] and [4,5] are considered overlapping.

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

Solution C++

// Author: https://blog.youkuaiyun.com/DarkRabbit
// Problem: https://leetcode.com/problems/merge-intervals/
// Difficulty: Medium
// Related Topics: `Array` `Sort`
// time: O(nlogn)

#pragma once

#include "pch.h"

namespace P56MergeIntervals
{
    class Solution
    {
        public:
        vector<vector<int>> merge(vector<vector<int>>& intervals)
        {
            vector<vector<int>> results;

            sort(intervals.begin(), intervals.end(), Compare);
            
            for (int i = 0; i < intervals.size(); i++)
            {
                if (results.empty() || results[results.size() - 1][1] < intervals[i][0])
                {
                    results.push_back(intervals[i]);
                }
                else if (results[results.size() - 1][1] < intervals[i][1])
                {
                    results[results.size() - 1][1] = intervals[i][1];
                }
            }

            return results;
        }

        static bool Compare(vector<int>& left, vector<int>& right)
        {
            return left[0] < right[0];
        }
    };
}