[Leetcode 001] Two sum

Question:

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

You may assume that each input wocould haveExactlyOne solution.

Example:

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

Method:

Java

• 1. Dual-cycle of violence, time complexity is O (n ^ 2)

1. publicclassSolution{
2. publicint[] twoSum(int[] nums,int target){
3. int[] arr =newint[2];
4. for(int i =0;i< nums.length-1; i++){
5. for(int j = i+1; j< nums.length; j++){
6. if(target == nums[i]+ nums[j]){
7. arr[0]= i ;
8. arr[1]= j ;
9. }
10. }
11. }
12. return arr ;
13. }
14. }

• 2. With hashmap or hashtable, the time complexity is O (n)

1. publicclassSolution{
2. publicint[] twoSum(int[] nums,int target){
3. int[] result ={0,0};
4. Map<Integer,Integer> map =newHashMap<Integer,Integer>();
5. for(int i =0;i<nums.length;i++){
6. if(map.containsKey(target-nums[i])){
7. result[0]= map.get(target-nums[i]);
8. result[1]= i;
9. }else{
10. map.put(nums[i],i);
11. }
12. }
13. return result ;
14. }
15. }

Analysis of ideas:
Obviously better than the first solution, greatly reducing the time complexity
Create a hash table and use the array's V value and index value as the K-V of the hash table
When target-K exists in the table, the corresponding value is taken out, which is the location to be searched.

Python
The same hashtable idea, key is number, value is Index

1. classSolution(object):
2. def twoSum(self, nums, target):
3. """
4. :type nums: List[int]
5. :type target: int
6. :rtype: List[int]
7. """
8. index ={};
9. for i,j in enumerate(nums):
10. if target-j in index:
11. return[index[target-j],i]
12. index[j]= i

From Weizhi note (wiz)

[Leetcode 001] Two sum