LeetCode算法(Python)--1、Two Sum

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Given an array of integers, return indices of the two numbers such that they add up to a specific target.

You may assume that each input would have exactly one solution, and you may not use the same element twice.

Example:

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Given nums = [2, 7, 11, 15], target = 9,
Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

这种蛮力方法是：

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class Solution(object):
def twoSum(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: List[int]
"""
for i in range(len(nums)):
for j in range(i+1, len(nums)):
if(nums[i] + nums[j] == target):
return [i, j]

另外一种比较好的做法是:

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class Solution(object):
def twoSum(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: List[int]
"""
arr = {};//使用字典存储遍历过的num和对应下标{num:index}
length = len(nums);//计算输入的列表长度
for i in range(length):
if (target - nums[i]) in arr://如果target-当前num的差在arr中，则表示已经找到答案，返回结果即可
return [arr[target - nums[i]], i];
arr[nums[i]] = i;//否则，将该num及其下标存入arr中