[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：

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：

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:

"""
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++：

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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