【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;}}