57.Insert Interval¶

57.Insert Interval¶

Tags: Hard Array Sort

Links: https://leetcode.com/problems/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].

NOTE: input types have been changed on April 15, 2019. Please reset to default code definition to get new method signature.

这道题目很容易就联想到56题，把newInterval插入到intervals里排个序，剩下就方法同理了。

class Solution {
public:
    vector<vector<int>> insert(vector<vector<int>>& intervals, vector<int>& newInterval) {
        intervals.push_back(newInterval);
        sort(intervals.begin(), intervals.end());
        vector<vector<int>> res;
        res.push_back(intervals[0]);

        for (int i = 1; i < intervals.size(); ++i) {
            if (intervals[i][0] > res.back()[1]) {
                res.push_back(intervals[i]);
            }
            else {
                res.back()[1] = max(intervals[i][1], res.back()[1]);
            }
        }

        return res;
    }
};

Runtime: 28 ms, faster than 15.95% of C++ online submissions for Insert Interval.
Memory Usage: 12.6 MB, less than 40.00% of C++ online submissions for Insert Interval.

从运行的结果来看，算法不太理想，因为注意到这个题和原来的问题有一个比较明显的区别，就是数组本来就是排好序的，所以应该充分利用这一点。

class Solution {
public:
    vector<vector<int>> insert(vector<vector<int>>& intervals, vector<int>& newInterval) {
        vector<vector<int>> res;
        int cur = 0, n = intervals.size();
        while (cur < n && newInterval[0] > intervals[cur][1])
            res.push_back(intervals[cur++]);

        while (cur < n && intervals[cur][0] <= newInterval[1]) {
            newInterval[0] = min(newInterval[0], intervals[cur][0]);
            newInterval[1] = max(newInterval[1], intervals[cur][1]);
            ++cur;
        }
        res.push_back(newInterval);

        while (cur < n) res.push_back(intervals[cur++]);

        return res;
    }
};

处理起来主要分为三段进行处理，因为区间序列有序且不重叠，所以需要去界定newInterval的左端点在什么位置，很显然，如果当前区间的右端点小于新区间的左端点，那么必然没有重叠的部分。所以直接放入结果数组即可。

如果新区间和当前区间产生了重叠，但是并不清楚它覆盖了多大的范围，采取的策略就是逐个探测并合并。什么情况停止合并，显然是没有和newInterval重叠的时候，那么就是新区间的右端点小于intervals的左端点，所以余下的部分直接推入结果数组即可。

进一步思考：因为题目里限定了区间是没有重叠的，如果把这个条件去掉，但是区间仍然是有序的，那么就需要先进行一次区间合并，然后再按照上面的算法遍历一便，时间复杂度不会改变，仍然是 $O(n)$ 。