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【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].
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：
1 /** 2 * Definition for an interval. 3 * public class Interval { 4 * int start; 5 * int end; 6 * Interval() { start = 0; end = 0; } 7 * Interval(int s, int e) { start = s; end = e; } 8 * } 9 */10 public class Solution {11 public List<Interval> insert(List<Interval> intervals, Interval newInterval) {12 List<Interval> answer = new ArrayList<Interval>();13 int insertPos = 0;14 for(Interval in:intervals){15 if(newInterval.start > in.end){16 answer.add(in);17 insertPos++;18 }19 else if(in.start > newInterval.end)20 {21 answer.add(in);22 }23 else{24 newInterval.start = Math.min(in.start, newInterval.start);25 newInterval.end = Math.max(in.end, newInterval.end);26 } 27 }28 answer.add(insertPos, newInterval);29 return answer;30 }31 }
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