Leetcode刷题笔记（持续更新...）

标签：Leetcode   算法   Python   

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

Example:
Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

暴力算法时间复杂度为O(n2)，不可取

方法一：

时间复杂度O(n)算法：

class Solution:
    def twoSum(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[int]
        """
        dic = {}
        temp = 0
        for i,j in enumerate(nums): #i存储nums的索引
            temp = target - j
            if j in dic:
                return [dic[j],i]
            dic[temp] = i #dic的值存储nums的索引，键存储该索引对应的nums值与target的差

Leetcode刷题笔记（持续更新...）

标签：Leetcode   算法   Python   

原文地址：http://blog.51cto.com/11281400/2103588