用unordered_map来做，扫描一遍数组即可，时间复杂度O(n)
class Solution { public: vector<int> twoSum(vector<int>& nums, int target) { unordered_map<int, int> heap; for(int i = 0; i < nums.size(); i ++) { if(heap.count(target - nums[i])) return {heap[target - nums[i]], i}; heap[nums[i]] = i; } return {}; //避免编译出错 } };