19.2.9 [LeetCode 57] Insert Interval

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:

Input: intervals = [[1,3],[6,9]], newInterval = [2,5]
Output: [[1,5],[6,9]]

Example 2:

Input: intervals = [[1,2],[3,5],[6,7],[8,10],[12,16]], newInterval = [4,8]
Output: [[1,2],[3,10],[12,16]]
Explanation: Because the new interval [4,8] overlaps with [3,5],[6,7],[8,10].

题意

给定一组以左边界大小排列的互不重合的闭区间，求加入一个新区间后的集合（也需要从小到大排列）

题解

一开始做的十分复杂

class Solution {
public:
    vector<Interval> insert(vector<Interval>& intervals, Interval newInterval) {
        if (intervals.size() < 1) {
            intervals.push_back(newInterval);
            return intervals;
        }
        int s = newInterval.start, e = newInterval.end, i;
        for (i = 0; i < intervals.size(); i++) {
            if (intervals[i].start <= s && intervals[i].end >= s) {
                if (e <= intervals[i].end)
                    return intervals;
                break;
            }
            else if (intervals[i].start > s) {
                if(intervals[i].start>e) {
                    intervals.insert(intervals.begin()+i,newInterval);
                    return intervals;
                }
                intervals[i].start = s;
                if (e <= intervals[i].end)
                    return intervals;
                break;
            }
        }
        int mark = i;
        i++;
        for (; i < intervals.size(); ) {
            if (intervals[i].start > e) {
                intervals[mark].end = e;
                return intervals;
            }
            else if (intervals[i].end >= e) {
                intervals[mark].end = intervals[i].end;
                intervals.erase(intervals.begin() + i);
                return intervals;
            }
            intervals.erase(intervals.begin() + i);
        }
        if(mark>=intervals.size())
            intervals.push_back(newInterval);
        else
            intervals[mark].end = e;
        return intervals;
    }
};

需要考虑很多条件，于是重写

class Solution {
public:
    vector<Interval> insert(vector<Interval>& intervals, Interval newInterval) {
        int i = 0, s = newInterval.start, e = newInterval.end;
        vector<Interval>ans;
        while (i < intervals.size() && intervals[i].end < s) {
            ans.push_back(intervals[i]);
            i++;
        }
        int left = s, right = e;
        while (i < intervals.size() && intervals[i].start <= e) {
            left = min(left, intervals[i].start);
            right = max(right, intervals[i].end);
            i++;
        }
        ans.push_back(Interval(left, right));
        while (i < intervals.size()) {
            ans.push_back(intervals[i]);
            i++;
        }
        return ans;
    }
};

这个就快得多了……