Insert Interval

https://leetcode.com/problems/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].

解题思路：

这题和上一题 Merge Intervals 很像，区别就是1.原list已经根据start排序了，2.所有元素没有overlapping。反而比较简单。

1. 从头至尾遍历list，找到start比newInterval的start大的，就在它前面插入newInterval。同时跳出遍历。

2. 如果已经遍历结束，就在结尾插入newInterval。

3. 处理newInterval前面节点overlap的问题。如果newInterval不是第一个节点，看它的start和前一个节点的end，小于或等于end，就用他俩end里较大的值去更新前一个元素的end，同时将newInterval移除。同时将newInterval的前一个元素视为newInterval。

4. 处理newInterval和后面节点的overlap问题。如果newInterval的end>=后一个元素的start，用两者end的较大值去更新newInterval，同时移除后一个节点。否则跳出循环，或一直循环处理直到newInterval为最后一个节点。

/**
 * 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 List<Interval> insert(List<Interval> intervals, Interval newInterval) {
        if(intervals.size() == 0) {
            intervals.add(newInterval);
            return intervals;
        }
        int insertIndex = 0;
        while(insertIndex < intervals.size()) {
            if(intervals.get(insertIndex).start > newInterval.start) {
                intervals.add(insertIndex, newInterval);
                break;
            }
            insertIndex++;
        }
        if(insertIndex == intervals.size()) {
            intervals.add(newInterval);
        }
        
        if(insertIndex - 1 >= 0) {
            if(intervals.get(insertIndex - 1).end >= newInterval.start) {
                intervals.get(insertIndex - 1).end = Math.max(intervals.get(insertIndex - 1).end, intervals.get(insertIndex).end);
                intervals.remove(insertIndex);
                insertIndex--;
            }
        }
        
        while(insertIndex < intervals.size() - 1 && intervals.get(insertIndex).end >= intervals.get(insertIndex + 1).start) {
            intervals.get(insertIndex).end = Math.max(intervals.get(insertIndex).end, intervals.get(insertIndex + 1).end);
            intervals.remove(insertIndex + 1);
        }
        return intervals;
    }
}

其实插入后，直接调用 Merge Intervals 题目里的方法也可以。

/**
 * 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 class Compare implements Comparator<Interval> {
        public int compare(Interval i1, Interval i2) {
            return i1.start - i2.start;
        }
    }
    
    public List<Interval> insert(List<Interval> intervals, Interval newInterval) {
        intervals.add(newInterval);
        return merge(intervals);
    }
    
     public List<Interval> merge(List<Interval> intervals) {
        List<Interval> result = new ArrayList<Interval>();
        if(intervals == null || intervals.size() == 0 || intervals.size() == 1) {
            return intervals;
        }
        Collections.sort(intervals, new Compare());
        result.add(intervals.get(0));
        for(int i = 1; i < intervals.size(); i++) {
            Interval cur = intervals.get(i);
            Interval lastResult = result.get(result.size() - 1);
            if(lastResult.end >= cur.start) {
                lastResult.end = Math.max(lastResult.end, cur.end);
            } else {
                result.add(cur);
            }
        }
        return result;
    }
}

 //20180926

1. 将原来intervals中的interval插入结果，直到一个range的end大于等于newInterval的sart，说明有重叠。

2.将原来intervals从当前位置开始的所有range与newInterval合并，直到start大于newinterval的end，说明没有重叠了。注意的是要将合并后的range一次性插入。

3. 将剩下的range插入结果。

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