Intervals

【LeetCode】

区间。LeetCode第56题和57题。

56.  Insert Interval

下面这种解法是原地插入的方法，显然比新建一个列表省空间。

/* 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]. */


/**
 * 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) {
        int i = 0;
        while(i < intervals.size() && intervals.get(i).end < newInterval.start)
            i++;
        while(i < intervals.size() && intervals.get(i).start <= newInterval.end){
            newInterval = new Interval(Math.min(intervals.get(i).start, newInterval.start), Math.max(intervals.get(i).end, newInterval.end));
            intervals.remove(i);
        }
        intervals.add(i, newInterval);
        return intervals;
    }
}

57 Merge Intervals

三种解法，三种思路 ，第一种最快。

/* Given a collection of intervals, merge all overlapping intervals.

For example,
Given [1,3],[2,6],[8,10],[15,18],
return [1,6],[8,10],[15,18]. */

/**
 * 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; }
 * }
 */
 
//Solution1
// The idea is that for the result distinct Interval, the latter one's start must > previous one's end.
class Solution {
    public List<Interval> merge(List<Interval> intervals) {
        int n = intervals.size();
        int[] start = new int[n];
        int[] end = new int[n];
        for(int i = 0; i < n; i++){
            start[i] = intervals.get(i).start;
            end[i] = intervals.get(i).end;
        }
        Arrays.sort(start);
        Arrays.sort(end);
        List<Interval> res = new ArrayList<>();
        for(int i = 0, j = 0; i < n; i++) {
            if(i == n - 1 || start[i+1] > end[i]){
                res.add(new Interval(start[j], end[i]));
                j = i + 1;
            }
        }
        return res;
    }
}

//Solution2
//The idea is to sort the intervals by their starting points.
class Solution {
    public List<Interval> merge(List<Interval> intervals) {
        if(intervals.size() <= 1)
            return intervals;
        intervals.sort((i1, i2) -> Integer.compare(i1.start, i2.start));
        int start = intervals.get(0).start;
        int end = intervals.get(0).end;
        List<Interval> res = new ArrayList<>();
        for(Interval i : intervals){
            if(i.start <= end)
                end = Math.max(end, i.end); // attention this line, important
            else{
                res.add(new Interval(start, end));
                start = i.start;
                end = i.end;
            }
        }
        res.add(new Interval(start, end));
        return res;
    }
}

//Solution3
//use the Insert Interval code, construct a new list, and insert every interval into it
//to reuse the existing code or thoughts is very useful and should take care of it