[LeetCode] 56. Merge Intervals 合并区间

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 considerred overlapping.

给定一个有重叠的区间集合，合并所有重叠的区间。先对区间以第一个元素进行排序，定义一个变量result记录合并后的区间。然后迭代这些区间，如果区间的开始值大于result的最后一个区间的结尾值，说明整个区间都在result的最后一个区间的右侧，直接添加到result。如果区间开始值小于result的最后一个区间的结尾值，则有重合需要合并，result中的最后一个区间的尾部值变为两个尾部中的最大值。

Java：

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public class Solution {     public List<Interval> merge(List<Interval> intervals) {         if (intervals == null || intervals.size() <= 1) {             return intervals;         }                   Collections.sort(intervals, new IntervalComparator());                   List<Interval> result = new ArrayList<Interval>();         Interval last = intervals.get(0);         for (int i = 1; i < intervals.size(); i++) {             Interval curt = intervals.get(i);             if (curt.start <= last.end ){                 last.end = Math.max(last.end, curt.end);             }else{                 result.add(last);                 last = curt;             }         }                   result.add(last);         return result;     }                 private class IntervalComparator implements Comparator<Interval> {         public int compare(Interval a, Interval b) {             return a.start - b.start;         }     }   }

Java：

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class Solution {     /**      * @param intervals, a collection of intervals      * @return: A new sorted interval list.      */     public List<Interval> merge(List<Interval> intervals) {         List<Interval> ans = new ArrayList<>();           intervals.sort(Comparator.comparing(i -> i.start));           Interval last = null;         for (Interval item : intervals) {             if (last == null || last.end < item.start) {                 ans.add(item);                 last = item;             } else {                 last.end = Math.max(last.end, item.end);             }         }         return ans;     } }

Python:

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""" Definition of Interval. class Interval(object):     def __init__(self, start, end):         self.start = start         self.end = end """   class Solution:     # @param intervals, a list of Interval     # @return a list of Interval     def merge(self, intervals):         intervals = sorted(intervals, key=lambda x: x.start)         result = []         for interval in intervals:             if len(result) == 0 or result[-1].end < interval.start:                 result.append(interval)             else:                 result[-1].end = max(result[-1].end, interval.end)         return result

C++：

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class Solution { public:     static bool comp(const Interval &a, const Interval &b) {         return (a.start < b.start);     }     vector<Interval> merge(vector<Interval> &intervals) {         vector<Interval> res;         if (intervals.empty()) return res;         sort(intervals.begin(), intervals.end(), comp);         res.push_back(intervals[0]);         for (int i = 1; i < intervals.size(); ++i) {             if (res.back().end >= intervals[i].start) {                 res.back().end = max(res.back().end, intervals[i].end);             } else {                 res.push_back(intervals[i]);             }         }         return res;     } };

　　　　

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[LeetCode] 57. Insert Interval 插入区间

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