Leetcode | 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:
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].

思路就是找到要merge的n个区间，然后把这几个区间删掉，再把新的区间插进去。

s初始化为-1，s+1就是插入的位置。s记录的是要merge的区间的前一个区间，所以插入位置就是s+1。

i就是要merge的区间的后一个区间。

如果merge错的话，可能会出现Output Limit Exceeded。

56ms。

class Solution {
public:
    vector<Interval> insert(vector<Interval> &intervals, Interval newInterval) {
        int s = -1, i = 0;
        for (; i < intervals.size(); ++i) {
            if (newInterval.end < intervals[i].start) {
                break;
            }
            if (newInterval.start > intervals[i].end) {
                s = i;
                continue;
            }
            
            if (intervals[i].start < newInterval.start) newInterval.start = intervals[i].start;
            if (intervals[i].end > newInterval.end) newInterval.end = intervals[i].end;
        }
        
        if (i - s >= 2) intervals.erase(intervals.begin() + s + 1, intervals.begin() + i);
        intervals.insert(intervals.begin() + s + 1, newInterval);
        return intervals;
    }
};

 第三次写，用一个新的vector复制，因为vector的删除也是复制，这样还高效一点。

1. 不可能有交集的区间直接复制。

2. 有交集的区间求最终的交集。

3. 剩下的又直接复制。

这样可以不用对空输入进行特殊处理。

/**
 * 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> ans;
        
        int i = 0;
        for (; i < intervals.size(); ++i) {
            if (intervals[i].end < newInterval.start) {
                ans.push_back(intervals[i]);
            } else if (intervals[i].start > newInterval.end) {
                break;
            } else {
                if (intervals[i].end > newInterval.end) newInterval.end = intervals[i].end;
                if (intervals[i].start < newInterval.start) newInterval.start = intervals[i].start;
            }
        }
        ans.push_back(newInterval);
        for (;i < intervals.size(); ++i) {
            ans.push_back(intervals[i]);
        }
        
        return ans;
    }
};