【LeetCode】56. Merge Intervals 解题报告（Python & C++ & Java）

作者： 负雪明烛 id： fuxuemingzhu 个人博客： http://fuxuemingzhu.cn/

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目录

题目描述题目大意解题方法日期

题目地址：https://leetcode.com/problems/merge-intervals/#/description

题目描述

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.

题目大意

这个题目意思是在数轴上有多个区间，如果能合并成更大区间的就合并在一起。

解题方法

首先按照每个区间的start排序，然后遍历。

用start,end两个指针记录当前的区间的开始和结束，之后的工作就是比较每一个区间的开始是否小于等于上个区间的end值，如果小于等于说明有重叠、可以合并成更大区间，这个时候要选择当前的区间end和上个区间end的最大值作为新的end完成区间合并。如果当前区间的start大于上个区间的end，那么两个区间不能合并，故以start和end为开始和结束构建区间的放到结果list中。如此遍历所有，最后一个区间也要同样放到结果list中。

python代码如下：

# Definition for an interval. # class Interval(object): # def __init__(self, s=0, e=0): # self.start = s # self.end = e class Solution(object): def merge(self, intervals): """ :type intervals: List[Interval] :rtype: List[Interval] """ N = len(intervals) if not N: return [] intervals.sort(key = lambda x : x.start) res = [] start = intervals[0].start end = intervals[0].end for it in intervals: if it.start <= end: end = max(end, it.end) else: cur = Interval(start, end) res.append(cur) start = it.start end = it.end res.append(Interval(start, end)) return res

Java代码如下：

/** * Definition for an interval. * public class Interval { * int start; * int end; * Interval() { start = 0; end = 0; } * Interval(int s, int e) { start = s; end = e; } * } */ public class Solution { public List<Interval> merge(List<Interval> intervals) { if (intervals.size() <= 1) { return intervals; } Collections.sort(intervals, new Comparator<Interval>() { @Override public int compare(Interval o1, Interval o2) { return o1.start - o2.start; } }); int start = intervals.get(0).start; int end = intervals.get(0).end; List<Interval> answer = new ArrayList<Interval>(); for (Interval interval : intervals) { if (interval.start <= end) { end = Math.max(end, interval.end); } else { answer.add(new Interval(start, end)); start = interval.start; end = interval.end; } } answer.add(new Interval(start, end)); return answer; } }

C++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> merge(vector<Interval>& intervals) { const int N = intervals.size(); vector<Interval> res; if (N == 0) return res; sort(intervals.begin(), intervals.end(), [](Interval a, Interval b){ return a.start < b.start; }); int start = intervals[0].start; int end = intervals[0].end; for (auto& it : intervals) { if (it.start <= end) { end = max(it.end, end); } else { res.push_back(Interval(start, end)); start = it.start; end = it.end; } } res.push_back(Interval(start, end)); return res; } };

日期

2017 年 4 月 4 日 2019 年 1 月 9 日 —— 抓紧时间学习啊！