Leetcode: 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].

If output is List:

 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         if (intervals.size() == 0) {
13             intervals.add(newInterval);
14             return intervals;
15         }
16         ArrayList<Interval> res = new ArrayList<Interval>();
17         int cur = 0;
18         while (cur < intervals.size() && intervals.get(cur).end < newInterval.start) {
19             res.add(intervals.get(cur));
20             cur++;
21         }
22         while (cur < intervals.size() && intervals.get(cur).start <= newInterval.end) {
23             newInterval.start = Math.min(newInterval.start, intervals.get(cur).start);
24             newInterval.end = Math.max(newInterval.end, intervals.get(cur).end);
25             cur++;
26         }
27         res.add(newInterval);
28         while (cur < intervals.size()) {
29             res.add(intervals.get(cur));
30             cur++;
31         }
32         return res;
33     }
34 }

If output is array:

 1 class Solution {
 2     public int[][] insert(int[][] intervals, int[] newInterval) {
 3         int i = 0;
 4         ArrayList<int[]> list = new ArrayList<int[]>();
 5         while (i < intervals.length && intervals[i][1] < newInterval[0]) {
 6             list.add(intervals[i]);
 7             i ++;
 8         }
 9         
10         while (i < intervals.length && newInterval[1] >= intervals[i][0]) {
11             newInterval[0] = Math.min(newInterval[0], intervals[i][0]);
12             newInterval[1] = Math.max(newInterval[1], intervals[i][1]);
13             i ++;
14         }
15         
16         list.add(newInterval);
17         
18         while (i < intervals.length) {
19             list.add(intervals[i]);
20             i ++;
21         }
22         
23         int[][] res = new int[list.size()][2];
24         for (int j = 0; j < list.size(); j ++) {
25             res[j] = list.get(j);
26         }
27         return res;
28      }
29 }

24-27 line can also be

        int[][] res = new int[list.size()][2];
        res = list.toArray(res);
        return res;

posted @ 2014-10-05 12:33 neverlandly 阅读(413) 评论(0) 编辑收藏举报

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