- 题目2. 解答C++遍历数组，将数组中的元素和索引分别作为 unordered_map 的键和值，如果目标值和当前元素的差已经存于在map 的键中，即找到结果。class Solution {public: vector<int> twoSum(vector<int>& nums, int target) { unordered_map<int, int> tabel; vector<int> index; int n = nums.size(); for (int i = 0; i < n; i++) { int val = target - nums[i]; if (tabel.count(val) == 1) { index.push_back(tabel[val]); index.push_back(i); return index; } else { tabel[nums[i]] = i; } } }};Python遍历数组，将数组中的元素和索引分别作为字典的键和值，如果目标值和当前元素的差已经存于在字典的键中，即找到结果。class Solution: def twoSum(self, nums, target): """ :type nums: List[int] :type target: int :rtype: List[int] """ hash_table = {} for index, value in enumerate(nums): num = target - value if num in hash_table: return [hash_table[num], index] hash_table[value] = index获取更多精彩，请关注「seniusen」!