Two Sum

Two Sum

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

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].

开始做没有别的方法，就是暴力解决：

    public int[] twoSum(int[] nums, int target) {             int []res = new int[2];            for(int i=0;i<nums.length;i++){                int exp = target-nums[i];                for(int j = i+1;j<nums.length;j++){                if(nums[j]==exp){                    res[0] = i;                    res[1] = j;                    break A;                }              }           }            return res;    }}

这样的话时间复杂度O(n²),空间复杂度O(1)，
对于大数组的话运行时间比较大。
查询解决方法后，用hash表来存储，牺牲空间节省时间时间复杂度O(n),空间也是O(n)，
代码如下：

public class Solution {    public int[] twoSum(int[] nums, int target) {             int []res = new int[2];                Map<Integer,Integer> map = new HashMap<>(nums.length);            for(int i=0;i<nums.length;i++){                Integer exp = target-nums[i];                if(map.containsKey(exp)){                    res[0]= map.get(exp);                    res[1] = i;                    break;                }                map.put(nums[i],i);           }            return res;    }}