Waiter finally opened a blog, really happy. Recently looking at Java, so take leetcode practice practiced hand. In the future I will be in the process of the dissolution of their thoughts to write down, to share to everyone, which has my own original part, but also to refer to other people's code to join their own understanding of the part, I hope you see more comments, a refueling.
Problem Description:
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, target=9
Output: index1=1, index2=2
The idea of disintegration:
Set up a hashmap, where the key value key is the numeric value in the array, and the corresponding value is the subscript value of the set.
Iterate from the beginning of the array, and find out whether the value corresponding to Target-numbers[i] is present in HashMap each time. If the value already exists, then a value of two is found and the target value is returned, and if the value does not exist, the (Numbers[i], i) is placed in the HashMap.
The time complexity of the entire algorithm is O (n).
The code is as follows:
1 Public classSolution {2 Public int[] Twosum (int[] numbers,inttarget) { 3Map<integer, integer> map =NewHashmap<integer, Integer> (Numbers.length * 2);4 int[] results =New int[2];5 6 for(inti = 0; i < numbers.length; i++){7Integer Temp1 = Map.get (Target-numbers[i]);8 if(Temp1 = =NULL){9 Map.put (Numbers[i], i);Ten } One Else{ AResults[0] = i + 1; -RESULTS[1] = Temp1 + 1; - if(Results[0] > Results[1]){ the intTemp2 = Results[0]; -Results[0] = results[1]; -RESULTS[1] =Temp2; - } + returnresults; - } + } A return NULL; at } -}
Java [Leetcode 1]-Sum