Insert Interval

57. Insert Interval

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].

区间合并，并且输入的区间是不重叠的

Solution

两个区间i,j的位置有 ss si sb ii ib bb,水题

/**
 * Definition for an interval.
 * public class Interval {
 *     int start;
 *     int end;
 *     Interval() { start = 0; end = 0; }
 *     Interval(int s, int e) { start = s; end = e; }
 * }
 */
class Solution {
    public List<Interval> insert(List<Interval> intervals, Interval newInterval) {
        List<Interval> list = new ArrayList<>();
        boolean flag = true;
        for(Interval i: intervals){
            // s i
            if(i.start<newInterval.start && i.end>=newInterval.start && i.end<=newInterval.end){
                newInterval.start = i.start;
            }
            
            // i b
            if(i.start>=newInterval.start&& i.start<=newInterval.end && i.end>newInterval.end){
                newInterval.end = i.end;
            }
            
            // s b
            if(i.start<newInterval.start && i.end>newInterval.end){
                flag = false;
                list.add(i);
            }
            // i i

            // s s
            if(i.end<newInterval.start){
                list.add(i);
            }
            
            // b b            
            if(i.start>newInterval.end){
                if(flag==true){
                    list.add(newInterval);
                    flag=false;
                }
                list.add(i);
            }
        }
        if(flag==true){
            list.add(newInterval);
        }
        return list;
    }
}