update 763

Author: KaidiGuo

README.md

@@ -46,5 +46,7 @@ For question details, please go to [Leetcode](https://leetcode.com/problemset/al
 6. [Tree](https://github.com/KaidiGuo/Algorithm-Exercises/tree/master/Linked%20List)
    + Medium 105 Construct Binary Tree from Preorder and Inorder Traversal  -- recursive
 
+7. [Two Pointers](https://github.com/KaidiGuo/Algorithm-Exercises/tree/master/Linked%20List)
+   + Medium 763 Partition Labels
 
 

Two Pointers/763. Partition Labels.py

@@ -0,0 +1,18 @@
+class Solution:
+    def partitionLabels(self, s: str) -> List[int]:
+        p1,p2=0,0
+        most_right = { c : i for i,c in enumerate(s)}
+        result=[]
+        for i,c in enumerate(s):
+            p2 = max(p2, most_right[c])
+            if i==p2:
+                result.append(p2-p1+1)
+                p1=i+1
+        return result
+
+# Create a dictionary, keep track of the right most index of each character
+# Create two pointers, p1 and p2, starting from 0
+# Loop through S
+# Inside the loop, keep p2 always at the right most index of all looped characters (using max(p2,most_right{i}))
+# When i fininally catchs up p2, means all of the looped characters, their last appearences are smaller than p2 
+# You find the first segmentation/partition, now move p1 to i+1 for nect segmentation/partition