1. Two Sum (map ) - 爱码网

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].

 1 class Solution {
 2 public:
 3     vector<int> twoSum(vector<int>& nums, int target) {
 4        unordered_map<int,int> map;
 5        for(int i = 0; i < nums.size(); ++i) {
 6            auto iter = map.find(nums[i]);
 7            if (iter != map.end()) {
 8                return {map[nums[i]],i};
 9            } else {
10                 map[target - nums[i]] = i;
11            }
12        }
13        return {0,0};
14     }
15 };





法1 ：暴力

 1 class Solution {
 2     public int[] twoSum(int[] nums, int target) {
 3         int looks ;
 4         int[] res ={0,0};
 5         for(int i = 0;i<nums.length;i++){
 6             looks = target - nums[i];
 7             for(int j = i+1;j<nums.length;j++){
 8                 if((looks == nums[j])) {
 9                   res[0] = i;
10                   res[1] = j;
11                 } 
12             }
13         }
14         return res;
15     }
16 }

法2：利用dict 映射，
遍历一遍list,将target减去当前元素的差作为key，当前元素的下标为val
如果当前元素是字典的key ,则返回当前元素的下标，与字典中当前元素对应的val

 1 class Solution:
 2     def twoSum(self, nums, target):
 3         """
 4         :type nums: List[int]
 5         :type target: int
 6         :rtype: List[int]
 7         """
 8         n_dict = {}
 9         for i in range(len(nums)):
10             if nums[i] in n_dict.keys():
11                 return [n_dict[nums[i]],i]
12             else:
13                 n_dict[target-nums[i]] = i;
14

20180325
用 key in dict: 判断效率更高！！！
时间复杂度 o(n)
空间复杂度 o(n)

 1 class Solution:
 2     def twoSum(self, nums, target):
 3         """
 4         :type nums: List[int]
 5         :type target: int
 6         :rtype: List[int]
 7         """
 8         n_dict = {}
 9         for i,item in enumerate(nums):
10             val = target - item
11             if item in n_dict:
12                 return [n_dict[item],i]
13             else:
14                 n_dict[val] = i