create: 模板题2 - Insert Interval

Author: LLancelot

merge-intervals.md

@@ -1,6 +1,6 @@
 # Merge Intervals 区间合并问题
 
-## 模板题1 - Merge Intevals
+## 模板题1 - Merge Intervals
 
 #### 题目
 
@@ -64,3 +64,77 @@ class Solution(object):
         return merged_intervals
 ```
 
+## 模板题2 - Insert Interval
+
+#### 题目
+
+Given a list of non-overlapping intervals **sorted** by their start time, **insert a given interval at the correct position** and merge all necessary intervals to produce a list that has only mutually exclusive intervals.
+
+**Example 1:**
+
+```
+Input: Intervals=[[1,3], [5,7], [8,12]], New Interval=[4,6]
+Output: [[1,3], [4,7], [8,12]]
+Explanation: After insertion, since [4,6] overlaps with [5,7], we merged them into one [4,7].
+```
+
+**Example 2:**
+
+```
+Input: Intervals=[[1,3], [5,7], [8,12]], New Interval=[4,10]
+Output: [[1,3], [4,12]]
+Explanation: After insertion, since [4,10] overlaps with [5,7] & [8,12], we merged them into [4,12].
+```
+
+**Example 3:**
+
+```
+Input: Intervals=[[2,3],[5,7]], New Interval=[1,4]
+Output: [[1,4], [5,7]]
+Explanation: After insertion, since [1,4] overlaps with [2,3], we merged them into one [1,4].
+ 
+```
+
+总结：对于一个**排好序**的区间序列，插入一个新区间，返回插入且合并后的区间结果。
+
+#### 思路
+
+- 如果用上一道题 Merge Intervals 的方法，先排序，再合并，时间复杂度为 O(N*logN)。但考虑到本题所给区间序列已排好序，我们尝试用 O(N) 的办法一趟完成合并。
+
+#### 图解&算法
+
+![merge-intervals-2.png](https://github.com/LLancelot/LeetCode/blob/master/images/merge-intervals-2.png?raw=true)
+
+- 对第一种情况：也就是 a.end < b.start 的区间，将这些区间直接加入到 result 并跳过即可，直到发现有 overlap 的区间为止。
+- 对剩余四种情况，共同特点 a.start < b.end，即必有交集，需要合并：
+  - 合并后的起点 start，谁在前，谁是 start，即 ***min***(a.start, b.start)
+  - 合并后的终点 end，谁在后，谁是 end，即 ***max***(a.end, b.end)
+
+#### 代码
+
+```python
+class Solution(object):
+    def insert(self, intervals: List[List[int]], newInterval: List[int]) -> List[List[int]]:
+        
+        merged, i, start, end = [], 0, 0, 1
+        while i < len(intervals) and intervals[i][end] < newInterval[start]:
+            # skip these intervals and append them into merged[]
+            merged.append(intervals[i])
+            i += 1
+        
+        # already find interval do not follow the condition
+        while i < len(intervals) and intervals[i][start] <= newInterval[end]:
+            newInterval[start] = min(newInterval[start], intervals[i][start])	# 谁小谁起点
+            newInterval[end] = max(newInterval[end], intervals[i][end])			# 谁大谁终点
+            i += 1
+            
+        merged.append(newInterval)	# 合并完，加到merged
+        
+        # for the rest of intervals
+        while i < len(intervals):
+            merged.append(intervals[i])
+            i += 1
+        
+        return merged
+```
+