LeetCode_1-- two numbers (Two sum)

And programmers code farmers know the difference lies in data structures and algorithms, so I began to brush Leetcode topic of road, hee hee

topic

Given an array of integers, returns two numbers index , so that they are applied to specific targets.

You can assume that each input is only one solution, and you may not be twice using the same elements.

Given nums = [2,7,11,15], target = 9, 
since the nums [ 0 ] + the nums [ 1 ] = 2. 7 = +. 9, 
returns to [ 0 , 1 ].

A Solution

thought

 　　Brute force, twice the number of loops through the array is equal to the sum of each target, and then returns the index

class Solution {
    public int[] twoSum(int[] nums, int target) {
        int[] arrs=new int[2]; ;
        int sum=0;
        for (int i = 0; i <nums.length; i++) {
            for (int j = 0; j <nums.length&&i!=j ; j++) {
                sum=nums[i]+nums[j];
                if(target==sum){
                    arrs[0]=i;
                    arrs[1]=j;
                }

            }
        }

        return arrs;
    }
}

result

　　When execution: 95 MS , beat the 8.21% of all users to submit in Java　　

　　Memory consumption: 37.2 MB , defeated 90.74% of all users to submit in Java

　　When used too long, the typical time for space approach

Solution two

Thinking

int[] arrs=new int[2]; ;
        int sum=0;
        HashMap<Integer,Integer> hashMap=new HashMap<>();
        for (int i = 0; i <nums.length; i++) {
            hashMap.put(nums[i],i);
        }
        for (int i = 0; i <nums.length; i++) {
            int num=target-nums[i];
            if(hashMap.containsKey(num)&&hashMap.get(num)!=i){
                arrs[0]=i;
                arrs[1]=hashMap.get(num);
                break;
            }
        }
return arrs;

result

　　When execution: 7 ms, beat the 88.43% of all users to submit in Java

　　Memory consumption: 38.9 MB, defeated 51.18% of all users to submit in Java