search-in-rotated-sorted-array-ii.py

Author: Chris Wu

problems/search-in-rotated-sorted-array-ii.py

@@ -0,0 +1,37 @@
+"""
+We use binary search if the interval is sorted (`nums[l]<nums[r]`).
+If it is not sorted, we cut the array into half, to see
+    * If the left half is sorted and the target is in the range. If so, look into the left part. (`r = p-1`)
+    * If the right half is sorted and the target is in the range. If so, look into the right part. (`l = p+1`)
+    * Else we can only move the `r` and `l` to the middle. (`r = r-1`, `l = l+1`). For example, `[1,1,1,1,1,1,1,3,1,1,1]`.
+The time complexity sits between, `O(LogN)` and `O(N)`.
+If there are a lots of the same value, we like `[1,1,1,1,1,1,1,3,1,1,1]` the time will closer to O(N). Vise versa.
+"""
+class Solution(object):
+    def search(self, nums, t):
+        if nums is None or len(nums)==0: return False
+        l = 0
+        r = len(nums)-1
+
+        while l<=r:
+            p = (l+r)/2
+            if nums[l]==t or nums[p]==t or nums[r]==t:
+                return True
+
+            if nums[l]<nums[r]:
+                #check if out of ragne
+                if t<nums[l] or nums[r]<t: return False
+                #binary search
+                if t<nums[p]:
+                    r = p-1
+                else:
+                    l = p+1
+            else:
+                if nums[l]<nums[p] and nums[l]<t and t<nums[p]:
+                    r = p-1
+                elif nums[p]<nums[r] and nums[p]<t and t<nums[r]:
+                    l = p+1
+                else:
+                    r = r-1
+                    l = l+1
+        return False

problems/search-in-rotated-sorted-array.py

@@ -131,4 +131,3 @@ def search(self, nums, target):
 
 
 
-