【leetcode刷题笔记】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].

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题解：利用变量insertPos记录最终区间插入的位置。遍历intervals，那么当前遍历的interval和newInterval有以下几种情况：

代码如下：

/**
 * 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; }
 * }
 */
public class Solution {
    public List<Interval> insert(List<Interval> intervals, Interval newInterval) {
        List<Interval> answer = new ArrayList<Interval>();
        int insertPos = 0;
        for(Interval in:intervals){
            if(newInterval.start > in.end){
                answer.add(in);
                insertPos++;
            }
            else if(in.start > newInterval.end)
            {
                answer.add(in);
            }
            else{
                newInterval.start = Math.min(in.start, newInterval.start);
                newInterval.end = Math.max(in.end, newInterval.end);
            }            
        }
        answer.add(insertPos, newInterval);
        return answer;
    }
}