LeetCode 57 Insert Interval (C++)

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Given a set of non-overlapping intervals, insert a new interval into the intervals (merge if necessary).
You may assume that the intervals were initially sorted according to their start times.
Example 1:
Given intervals [1,3],[6,9], insert and merge [2,5] in as [1,5],[6,9].
Example 2:
Given [1,2],[3,5],[6,7],[8,10],[12,16], insert and merge [4,9] in as [1,2],[3,10],[12,16].
This is because the new interval [4,9] overlaps with [3,5],[6,7],[8,10].

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> insert(vector<Interval> &intervals, Interval newInterval) {
        vector<Interval> ret;
        int i = 0;
        for (i = 0; i < intervals.size(); ++i) {
            if (newInterval.start <= intervals[i].end) break;
            ret.push_back(intervals[i]);
        }
        if (i == intervals.size()) {
            ret.push_back(newInterval);
            return ret;
        }
        if (newInterval.end < intervals[i].start) {
            ret.push_back(newInterval);
        } else {
            int start = min(newInterval.start, intervals[i].start);
            int end = max(newInterval.end, intervals[i].end);
            while (++i < intervals.size() && intervals[i].start <= end) {
                end = max(end, intervals[i].end);
            }
            ret.push_back(Interval(start, end));
        }
        for (; i < intervals.size(); ++i) {
            ret.push_back(intervals[i]);   
        }
        return ret;
    }
};

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