【LeetCode】Merge Intervals && Insert Interval

1、Merge Intervals  
Total Accepted: 6989 Total Submissions: 34958 My Submissions
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].
注意给的区间可能是无序的，先排序后处理。
基本的思路还是比较合并吧，比较考察思维逻辑，注意边界处理。

Java AC

/**
 * 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 ArrayList<Interval> merge(ArrayList<Interval> intervals) {
        if(intervals == null){
            return intervals;
        }
        int size = intervals.size();
        if(size == 0 || size == 1){
            return intervals;
        }
        Collections.sort(intervals, new Comparator<Interval>() {

			public int compare(Interval o1, Interval o2) {
				if (o1.start == o2.start) {
					return o1.end - o2.end;
				}
				return o1.start - o2.start;
			}

		});
        ArrayList<Interval> list = new ArrayList<Interval>();
        Interval interval = intervals.get(0);
        int start = interval.start;
        int end = interval.end;
        for(int i = 1; i < size; i++){
            Interval intval = intervals.get(i);
            if(intval.start <= end){
            	int tempSize = list.size();
            	if (tempSize > 0) {
            		list.remove(tempSize-1);
				}
                start = Math.min(start, intval.start);
                end = Math.max(end, intval.end);
            }else{
            	if (i == 1) {
            		list.add(new Interval(start, end));
				}
                start = intval.start;
                end = intval.end;
            }
            list.add(new Interval(start, end));
        }
		return list;
    }
}

2、Insert Interval 

Total Accepted: 6691 Total Submissions: 33506 My Submissions
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].
求解给定的区间在已知区间的位置，将给定区间放入已知区间，然后就回到1的处理方式。

Java AC

/**
 * 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 ArrayList<Interval> insert(ArrayList<Interval> intervals,
			Interval newInterval) {
		ArrayList<Interval> list = new ArrayList<Interval>();
		if (intervals == null || intervals.size() == 0) {
			list.add(newInterval);
			return list;
		}
		int pos = searchInsert(intervals, newInterval.start);
		return merge(intervals, pos, newInterval);
		
	}
	public ArrayList<Interval> merge(ArrayList<Interval> intervals, 
				int pos, Interval newInterval) {
        if(intervals == null){
            return intervals;
        }
        int size = intervals.size();
        ArrayList<Interval> newList = new ArrayList<Interval>();
        if (pos == 0) {
			newList.add(newInterval);
			newList.addAll(intervals);
		}else if (pos == size) {
			newList.addAll(intervals);
			newList.add(newInterval);
		}else {
			int i = 0;
			while (i < size) {
				if (i == pos) {
					newList.add(newInterval);
				}
				newList.add(intervals.get(i));
				i++;
			}
		}
        return merge(newList);
    }
	public ArrayList<Interval> merge(ArrayList<Interval> intervals) {
        int size = intervals.size();
        ArrayList<Interval> list = new ArrayList<Interval>();
        Interval interval = intervals.get(0);
        int start = interval.start;
        int end = interval.end;
        for(int i = 1; i < size; i++){
            Interval intval = intervals.get(i);
            if(intval.start <= end){
            	int tempSize = list.size();
            	if (tempSize > 0) {
            		list.remove(tempSize-1);
				}
                start = Math.min(start, intval.start);
                end = Math.max(end, intval.end);
            }else{
            	if (i == 1) {
            		list.add(new Interval(start, end));
				}
                start = intval.start;
                end = intval.end;
            }
            list.add(new Interval(start, end));
        }
		return list;
    }
	public int searchInsert(ArrayList<Interval> intervals, int target) {
		int size = intervals.size();
		if (target < intervals.get(0).start) {
			return 0;
		}
		if (target > intervals.get(size - 1).end) {
			return size;
		}
		int low = 0;
		int high = size - 1;
		int mid = 0;
		while (low <= high) {
			mid = (low + high) >> 1;
			if (intervals.get(mid).start > target) {
				high = mid - 1;
			} else if (intervals.get(mid).start < target) {
				low = mid + 1;
			} else {
				return mid;
			}
		}
		return low;
	}
}