leetcode57- Insert Interval- medium

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].

 

O(n)

独立方法1:

关键是要1.根据铰接窗口延展newInterval左右一点点 2.把延展后的newInterval插在正确的位置。

1.遍历元素，同时做两件事情 a.非铰接的就直接放入结果，同时用一个指针记录着该放前面的放多少个了来指示最后把newInterval插在上面位置。b.铰接的来试着延展newInterval

2.插入newInterval。

 

依赖上题方法2:

首先先按照当前newInterval的start找到在原列表里按顺序应该插入的位置，直接插进去，然后再做一次上面的merge。（因为省去了sort，所以时间复杂度还是O(n)） 

 

1.实现 elegant版

/**
 * 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) {
        
        if (newInterval == null) {
            return intervals;
        }
        List<Interval> result = new ArrayList<>();
        if (intervals == null) {
            result.add(newInterval);
            return result;
        }
        
        int insertIdx = 0;
        for (Interval interval : intervals) {
            if (interval.end < newInterval.start) {
                result.add(interval);
                insertIdx++;
            } else if (interval.start > newInterval.end) {
                result.add(interval);
            } else {
                newInterval.start = Math.min(newInterval.start, interval.start);
                newInterval.end = Math.max(newInterval.end, interval.end);
            }
        }
        result.add(insertIdx, newInterval);
        
        return result;
    }
}

 

2.实现ugly版的，遍历了两次，第一次专门拓展，第二次专门加组

/**
 * 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) {
        
        if (newInterval == null) {
            return intervals;
        }
        List<Interval> result = new ArrayList<>();
        if (intervals == null || intervals.size() == 0) {
            result.add(newInterval);
            return result;
        }
        
        for (int i = 0; i < intervals.size(); i++) {
            if (intervals.get(i).end >= newInterval.start) {
                newInterval.start = Math.min(newInterval.start, intervals.get(i).start);
            }
            if (intervals.get(i).start <= newInterval.end) {
                newInterval.end = Math.max(newInterval.end, intervals.get(i).end);
            }
        }
        
        int i;
        for (i = 0; i < intervals.size(); i++) {
            if (intervals.get(i). end < newInterval.start) {
                result.add(intervals.get(i));
            } else {
                break;
            }
        }
        result.add(newInterval);
        for ( ; i < intervals.size(); i++) {
            if (intervals.get(i).start <= newInterval.end) {
                continue;
            } else {
                result.add(intervals.get(i));
            }
        }
        
        return result;
        
    }
}