LeetCode 56 Merge Intervals (C++, Python)

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Given a collection of intervals, merge all overlapping intervals.
For example,
Given [1,3],[2,6],[8,10],[15,18],
return [1,6],[8,10],[15,18].

Solution C++


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/**
 * 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) {
        if (intervals.size() < 2) return intervals;
        vector<Interval> ret;
        sort(intervals.begin(), intervals.end(), [](const Interval& i1, const Interval& i2) {
            return i1.start < i2.start;
        });
        ret.push_back(intervals[0]);
        for (int i = 1; i < intervals.size(); ++i) {
            if (intervals[i].start <= ret.back().end) {
                ret.back().end = max(ret.back().end, intervals[i].end);        
            } else {
                ret.push_back(intervals[i]);
            }
        }
        return ret;
    }
};
Solution Python


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# Definition for an interval.
# class Interval:
#     def __init__(self, s=0, e=0):
#         self.start = s
#         self.end = e

class Solution:
    # @param intervals, a list of Interval
    # @return a list of Interval
    def merge(self, intervals):
        if len(intervals) < 2: return intervals
        intervals.sort(key = lambda x: x.start)
        ret = [intervals[0]]
        for i in range(1, len(intervals)):
            if intervals[i].start <= ret[-1].end:
                ret[-1].end = max(ret[-1].end, intervals[i].end)
            else:
                ret.append(intervals[i])
        return ret

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