Leetcode1-two Sums

Topic https://leetcode.com/problems/two-sum/

Given an array of integers, find the numbers such that they add up to a specific target number.

The function twosum should return indices of the numbers such that they add up to the target, where index1 must is Les S than Index2. Please note that your returned answers (both Index1 and INDEX2) is not zero-based.

You may assume this each input would has exactly one solution.

Input:numbers={2, 7, one, A, target=9
Output:index1=1, index2=2

Ideas

First consider the time Complexity O (n^2) loop traversal, Leetcode can not endure.

The second scenario is O (N), when iterating through an array, the value is key, the index is a dictionary of value, and then each time the target value minus the current value from the dictionary gets index,index less than I, the description exists in the dictionary.

Code O (n^2)

classSolution1:#@param {integer[]} nums    #@param {integer} target    #@return {integer[]}    deftwosum (self, Nums, target): RLT= []        if  notNums:returnRLT forIinchRange (len (nums)): forJinchRange (Len (nums[i+1:])):                ifNums[i] + nums[i+j] = =target:rlt.append (i) rlt.append (i+j)returnRlt

Code O (N)

classSolution:#@param {integer[]} nums    #@param {integer} target    #@return {integer[]}    deftwosum (self, Nums, target): RLT= []        if  notNums:returnRLT num_dict= {}         forIinchRange (len (nums)):ifNums[i] not inchNum_dict:num_dict[nums[i]]=IPrintNums[i],i J= Num_dict.get (Target-nums[i], 1)            ifJ < I andJ >-1: Rlt.append (J+1) rlt.append (i+1)                returnRLTreturnRlt

Refer to Address 1. http://blog.csdn.net/jiadebin890724/article/details/23305449

Leetcode1-two Sums