1.两数之和

方法一：暴力法
复杂度分析：

• 时间复杂度：O(n²)
• 空间复杂度：O(1)

public class SolutionOne {
    public int[] twoSum(int[] nums, int target){
        for( int i = 0; i< nums.length ; i++){
            for (int j = i+ 1; j < nums.length ; j++){
                if (nums[j] == target - nums[i]){
                    return new int[]{i ,j};
                }
            }
        }
        throw new IllegalArgumentException("No two sum solution");
    }
}

方法二：一遍哈希表

• 时间复杂度：O(n)
• 空间复杂度：O(n)

public class SolutionTwo {
    public  int[] twoSum(int[] nums, int target){
        Map<Integer,Integer> map = new HashMap<>();
        for(int i = 0; i < nums.length ; i++){
            int complement = target - nums[i];
            if(map.containsKey(complement)){
                return new int[]{map.get(complement),i};
            }
            map.put(nums[i],i);
        }
        throw  new IllegalArgumentException("No two sum solution");
    }
}