public class Solution {    public int[] TwoSum(int[] numbers, int target) {        Dictionary
     
       dic = new Dictionary
      
       ();            int count = numbers.Count();            List
       
        
         > list = new List
         
          
           >(); for (int i = 0; i < count; i++) { if (!dic.ContainsKey(numbers[i])) { dic.Add(numbers[i], i); list.Add(new KeyValuePair
           
            (numbers[i], i)); } } int[] ary = new int[2]; for (int i = 0; i < list.Count; i++) { for (int j = i; j < list.Count; j++) { var d1 = list[i]; var d2 = list[j]; if (d1.Key + d1.Key == target && d2.Value - d1.Value > 1) { ary[0] = d1.Value + 1; ary[1] = d1.Value + 2; return ary; } else if (i != j && d1.Key + d2.Key == target) { ary[0] = d1.Value + 1; ary[1] = d2.Value + 1; return ary; } } } return ary; }}
           
          
         
        
       
      
     

补充一个“双指针”思想的解决方案，使用python实现：

1 class Solution: 2     def twoSum(self, numbers: 'List[int]', target: 'int') -> 'List[int]': 3         i=0 4         j=len(numbers)-1 5         while numbers[i] + numbers[j] != target: 6             if numbers[i] + numbers[j]>target: 7                 j-=1 8             else: 9                 i+=110 11         return [i+1,j+1]