Leetcode 57. Insert Interval

57. Insert Interval

• Total Accepted: 64309
• Total Submissions: 260619
• Difficulty: Hard

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：区间划分。先找到新区间的左边界，再找到新区间的右边界。画图会更加清晰。

左边界：不满足intervals[i].end<newInterval.start意味着intervals[i-1].end<newInterval.start<=intervals[i].end。

右边界：满足intervals[j].start<newInterval.end时，不断合并区间，将intervals[j]和newInterval合并为新的newInterval，再尝试和intervals[j+1]合并，直到不能合并。因为intervals[j]和intervals[j+1]是区间独立的，所以intervals[j]被合并不会影响intervals[j+1]。

代码：

/**
 * 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) {
        int i,j;
        vector<Interval> res;
        for(i=0;i<intervals.size()&&intervals[i].end<newInterval.start;i++){//i
            res.push_back(intervals[i]);
        }
        for(j=i;j<intervals.size()&&intervals[j].start<=newInterval.end;j++){
            newInterval=Interval(min(intervals[j].start,newInterval.start),max(intervals[j].end,newInterval.end));
        }
        res.push_back(newInterval);
        for(;j<intervals.size();j++){
            res.push_back(intervals[j]);
        }
        return res;
    }
};