add solution 57-Insert-Interval

Author: andy6804tw

Blind Curated 75/README.md

@@ -36,6 +36,10 @@
 - Unique Paths
 - Jump Game
 #### **Graph**
+| No.   | Problem  | Category  | Difficulty | Solouion |
+|:--------:|:-----------:|:---------:|:---------:|:---------:|
+57.|[Insert Interval](https://leetcode.com/problems/insert-interval/) |Graph |Medium |  [Python](/Blind%20Curated%2075/python/57-Insert-Interval.md)
+
 - Clone Graph
 - Course Schedule
 - Pacific Atlantic Water Flow

Blind Curated 75/python/57-Insert-Interval.md

@@ -0,0 +1,61 @@
+# 57. Insert Interval (Python)
+## Problem
+You are given an array of non-overlapping intervals intervals where intervals[i] = [starti, endi] represent the start and the end of the ith interval and intervals is sorted in ascending order by starti. You are also given an interval newInterval = [start, end] that represents the start and end of another interval.
+
+Insert newInterval into intervals such that intervals is still sorted in ascending order by starti and intervals still does not have any overlapping intervals (merge overlapping intervals if necessary).
+
+Return intervals after the insertion.
+
+**Example 1:**
+```
+Input: intervals = [[1,3],[6,9]], newInterval = [2,5]
+Output: [[1,5],[6,9]]
+```
+
+**Example 2:**
+```
+Input: intervals = [[1,2],[3,5],[6,7],[8,10],[12,16]], newInterval = [4,8]
+Output: [[1,2],[3,10],[12,16]]
+Explanation: Because the new interval [4,8] overlaps with [3,5],[6,7],[8,10].
+```
+
+## Constraints:
+- 0 <= intervals.length <= 104
+- intervals[i].length == 2
+- 0 <= starti <= endi <= 105
+- intervals is sorted by starti in ascending order.
+- newInterval.length == 2
+- 0 <= start <= end <= 105
+
+## Solution 1.
+這一解法分成三步驟：
+
+1. 將小於 newInterval 的區間塞入 newList 直到 newInterval[0]>intervals[i][1]
+2. 將有覆蓋到 newInterval 的區間合併，同時找出 min start 和 max end
+3. 塞入剩下未涵蓋的區間
+
+- 串列處理、Graph
+- Run Time: 63 ms
+
+```py
+class Solution(object):
+    def insert(self, intervals, newInterval):
+        
+        newList=[]
+        i=0
+        # add all the intervals ending before newInterval starts
+        while i<len(intervals) and newInterval[0]>intervals[i][1]:
+            newList.append(intervals[i])
+            i+=1
+        # merge all overlapping intervals to one considering newInterval
+        while i<len(intervals) and newInterval[1]>=intervals[i][0]:
+            newInterval = [min(newInterval[0],intervals[i][0]), max(newInterval[1],intervals[i][1])]
+            i+=1
+        newList.append(newInterval) # add the union of intervals we got
+        # add all the rest
+        while i<len(intervals):
+            newList.append(intervals[i])
+            i+=1
+            
+        return newList
+```