完成leecode中的两个题目.

Author: ssjssh

README.md

@@ -95,7 +95,7 @@ algorithm
 * 在学习redis源码的过程中修改各个数据结构的实现,目的是使用更加精细的规则去提高数据结构的性能
 * 添加更多注释并且格式化代码,注释中注重于设计的思考
 * 添加算法导论中其他高级的数据结构
-* 计划完成一些在线的题库
+* 计划完成一些在线的题库,比如leecode和projecteuler
 
 
 

src/leecode/__init__.py

@@ -0,0 +1,3 @@
+#!/usr/bin/env python
+#-*- coding:UTF-8
+__author__ = 'shenshijun'

src/leecode/add_two_numbers.py

@@ -0,0 +1,75 @@
+#!/usr/bin/env python
+# -*- coding:UTF-8
+__author__ = 'shenshijun'
+"""
+You are given two linked lists representing two non-negative numbers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.
+
+Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
+Output: 7 -> 0 -> 8
+"""
+
+
+class ListNode(object):
+    def __init__(self, x):
+        self.val = x
+        self.next = None
+
+
+class Solution(object):
+    def __addUpDigit(self, node, up_digit):
+        last_node = None
+        while node is not None and up_digit > 0:
+            node.val += up_digit
+            if node.val >= 10:
+                node.val -= 10
+                up_digit = 1
+            else:
+                up_digit = 0
+            last_node = node
+            node = node.next
+        if up_digit > 0 and last_node:
+            last_node.next = ListNode(up_digit)
+
+    def addTwoNumbers(self, l1, l2):
+        """
+        :type l1: ListNode
+        :type l2: ListNode
+        :rtype: ListNode
+        """
+        cur_l1 = l1
+        cur_l2 = l2
+        last_l1 = None
+        last_up_digit = 0
+        while cur_l1 is not None and cur_l2 is not None:
+            this_val = cur_l1.val + cur_l2.val + last_up_digit
+            if this_val >= 10:
+                cur_l1.val = this_val - 10
+                last_up_digit = 1
+            else:
+                cur_l1.val = this_val
+                last_up_digit = 0
+            last_l1 = cur_l1
+            cur_l1 = cur_l1.next
+            cur_l2 = cur_l2.next
+
+        if cur_l1 is None and cur_l2 is None and last_up_digit > 0 and last_l1 is not None:
+            last_l1.next = ListNode(last_up_digit)
+        self.__addUpDigit(cur_l1, last_up_digit)
+        self.__addUpDigit(cur_l2, last_up_digit)
+
+        # 如果l1比较短,需要把l2长的部分合并到l1后面
+        if cur_l2 is not None:
+            if last_l1 is None:
+                return l2
+            else:
+                last_l1.next = cur_l2
+
+        return l1
+
+
+def main():
+    pass
+
+
+if __name__ == "__main__":
+    main()

src/leecode/two_sum.py

@@ -0,0 +1,68 @@
+#!/usr/bin/env python
+# -*- coding:UTF-8
+"""
+Given an array of integers, find two numbers such that they add up to a specific target number.
+The function twoSum should return indices of the two numbers such that they add up to the target,
+where index1 must be less than index2.Please note that your returned answers (both index1 and index2) are not zero-based.
+
+You may assume that each input would have exactly one solution.
+
+Input: numbers={2, 7, 11, 15}, target=9
+Output: index1=1, index2=2
+"""
+
+__author__ = 'shenshijun'
+
+"""
+本题有两个思路:
+一:  利用排序和二分查找
+先排序得到一个排序的数列,复杂度是O(nlgn),然后遍历数列,每遍历到一个元素的时候使用二分查询查找另外一个数字是否存在,复杂度仍然是O(nlgn)
+
+二: 使用hash表
+思路一为了快速查找而把数据都排了序,为了查找,显然可以使用hash表.而hash表是更高效的实现.一遍完成,因此复杂度是O(n)
+"""
+
+
+class Solution(object):
+    def twoSum(self, nums, target):
+        """
+        :type nums: List[int]
+        :type target: int
+        :rtype: List[int]
+        """
+        d = {}
+        for ind, num in enumerate(nums):
+            # 为了防止列表中出现重复的数,并且保证列表前面的index比后面的index小
+            if num in d:
+                if type(d[num]) is list:
+                    d[num].append(ind)
+                else:
+                    d[num] = [d[num], ind]
+            else:
+                d[num] = ind
+
+        for num, ind in d.iteritems():
+            # 注意:要同时考虑ind和d[other_part]都为list的情况
+            # 还要考虑,找到的index不能是同一个,并且不能重复
+            this_indexs = ind if type(ind) == list else [ind]
+            other_part = target - num
+            for this_index in this_indexs:
+                if other_part in d:
+                    if type(d[other_part]) == list:
+                        for other_index in d[other_part]:
+                            if other_index != this_index:
+                                return [min(other_index, this_index) + 1, max(other_index, this_index) + 1]
+                    elif d[other_part] != this_index:
+                        return [min(d[other_part], this_index) + 1, max(d[other_part], this_index) + 1]
+
+        return [-1, -1]
+
+
+def main():
+    print(Solution().twoSum([0, 4, 3, 0], 0))
+    print(Solution().twoSum([2, 7, 11, 15], 9))
+    print(Solution().twoSum([-3, 4, 3, 90], 0))
+
+
+if __name__ == "__main__":
+    main()