[LeetCode] 57. Insert Interval Java

Posted 2020-10-21 BrookLearnData

题目：

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].

题意及分析：给出一个按升序排序的非重叠区间序列，插入一个新区间，合并有重叠的区间。

代码：

/**
 * 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 {
    //给出一个按升序排序的非重叠区间序列，插入一个新区间，合并有重叠的区间
    public List<Interval> insert(List<Interval> intervals, Interval newInterval) {
       List<Interval> res = new ArrayList<>();
        int i=0;
        while(i<intervals.size() && intervals.get(i).end<newInterval.start){ //交叉区间之前的直接加入
            res.add(intervals.get(i++));
        }
        while(i<intervals.size() && intervals.get(i).start <= newInterval.end){     //有交叉区间
            newInterval = new Interval(
                    Math.min(intervals.get(i).start,newInterval.start),
                    Math.max(intervals.get(i).end,newInterval.end)
            );
            i++;
        }
        res.add(newInterval);
        while (i < intervals.size())        //重叠区间之后的也直接加入
            res.add(intervals.get(i++));
        return res;
    }
}