no message

Author: wuduhren

problems/python3/3sum.py

@@ -0,0 +1,40 @@
+"""
+Time: O(N^2)
+Space: O(1)
+
+First of all, we need to sort the list, so that it is easier to know where to move the pointers.
+(Keep in mind that sorting takes O(NLogN) of time, but since judging from the problem this solution might takes at least O(N^2), so it is ok to sort.)
+
+Now, we have 3 unique pointers at the list. i, j, k, where i<j<k.
+For a given i, set j at i+1 and k at the rightest.
+
+If the sum is too large, our only option is to move k left, so we can find smaller sum.
+If the sum is too small, our only option is to move j right, so we can find larger sum.
+If the sum is what we want, add it to the ans.
+
+[0] we need to skip the num that is the same as the last one to remove duplicate. This is actually the hardest part of the problem.
+[1] if nums[i] is larger than 0, we can stop, since nums[i] and nums[k] are surely larger than nums[i]
+"""
+class Solution:
+    def threeSum(self, nums: List[int]) -> List[List[int]]:
+        ans = []
+        nums.sort()
+        
+        for i in range(len(nums)):
+            if 0<i and nums[i]==nums[i-1]: continue #[0]
+            if nums[i]>0: break #[1]
+            
+            j, k = i+1, len(nums)-1
+            while j<k:
+                if nums[i]+nums[j]+nums[k]>0:
+                    k -= 1
+                elif nums[i]+nums[j]+nums[k]<0:
+                    j += 1
+                else:
+                    ans.append((nums[i], nums[j], nums[k]))
+                    while 0<k-1 and nums[k-1]==nums[k]: k -= 1 #[0]
+                    while j+1<len(nums) and nums[j+1]==nums[j]: j += 1 #[0]
+                    k -= 1
+                    j += 1
+                    
+        return ans