update

yuhuacao/ReverseLinkedList.py

@@ -0,0 +1,31 @@
+题目:
+给定一个链表，把这个链表反转过来。
+示例:
+输入: 1->2->3->4->5->NULL
+输出: 5->4->3->2->1->NULL
+解题思路：
+比较简单，对链表遍历，需保存上一个指针和下一个指针。在遍历过程中理清三者关系，注意不要互相覆盖即可。
+
+# Definition for singly-linked list.
+# class ListNode(object):
+#     def __init__(self, x):
+#         self.val = x
+#         self.next = None
+
+class Solution(object):
+    def reverseList(self, head):
+        """
+        :type head: ListNode
+        :rtype: ListNode
+        """
+        if head==None:
+            return []
+        last_temp = None
+        temp = head
+        while(temp!=None):
+            temp_next = temp.next
+            temp.next = last_temp
+            last_temp = temp
+            temp = temp_next
+
+        return last_temp

yuhuacao/readme.md

@@ -2,3 +2,5 @@
 
 1. Merge Sorted Array
 2. Maximum Length of Repeated Subarray
+3. Reverse Linked List
+4. Three Sum

yuhuacao/threeSum.py

@@ -0,0 +1,45 @@
+"""
+Given an array S of n integers, are there elements a, b, c in S such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.
+
+Note: The solution set must not contain duplicate triplets.
+
+For example, given array S = [-1, 0, 1, 2, -1, -4],
+
+A solution set is: [ [-1, 0, 1], [-1, -1, 2] ]
+"""
+"""
+解题思路：
+1，先将数组排序。
+2，排序后，可以按照TwoSum的思路来解题。怎么解呢？可以将num[i]的相反数即-num[i]作为target，然后从i+1到len(num)-1的数组元素中寻找两个数使它们的和为-num[i]就可以了。下标i的范围是从0到len(num)-3。
+3，这个过程要注意去重。
+4，时间复杂度为O(N^2)。
+"""
+
+
+class Solution(object):
+  def threeSum(self, nums):
+    """
+    :type nums: List[int]
+    :rtype: List[List[int]]
+    """
+    res = []
+    nums.sort()
+    for i in range(0, len(nums)):
+      if i > 0 and nums[i] == nums[i - 1]:
+        continue
+      target = 0 - nums[i]
+      start, end = i + 1, len(nums) - 1
+      while start < end:
+        if nums[start] + nums[end] > target:
+          end -= 1
+        elif nums[start] + nums[end] < target:
+          start += 1
+        else:
+          res.append((nums[i], nums[start], nums[end]))
+          end -= 1
+          start += 1
+          while start < end and nums[end] == nums[end + 1]:
+            end -= 1
+          while start < end and nums[start] == nums[start - 1]:
+            start += 1
+    return res