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leetcode : Two Sum

时间:2015-09-02 10:45:33      阅读:125      评论:0      收藏:0      [点我收藏+]

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题目: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

思路:开始拿到这一题以为很简单,就简单的两层遍历就好了,没想到超时了,好吧这又是一道拿空间换时间的算法题,我的解题思想就是拷贝到新的数组排序,从两头开始遍历。代码如下:

 1 class Solution(object):
 2     def twoSum(self, nums, target):
 3         """
 4         :type nums: List[int]
 5         :type target: int
 6         :rtype: List[int]
 7         """
 8         import copy
 9         
10         index = []
11         box = copy.deepcopy(nums)
12         
13         box.sort()
14         boxlen = len(box)
15         
16         i = 0
17         j = boxlen - 1
18         while i <= j:
19             if box[i] + box[j] == target:
20                 index.append(nums.index(box[i]) + 1)
21                 if box[i] == box[j]:
22                     index.append(nums.index(box[j],nums.index(box[i]) + 1) + 1)
23                 else:
24                     index.append(nums.index(box[j]) + 1)
25                 index.sort()
26                 return index
27             elif box[i] + box[j] < target:
28                 i += 1
29             else:
30                 j -=1
31                 
32         return index

 

leetcode : Two Sum

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原文地址:http://www.cnblogs.com/tingyu-yudeyinji/p/4777791.html

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