Description

LINK: https://leetcode.com/problems/two-sum

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:

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Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

Solution

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static auto x = [](){
    // turn off sync
    std::ios::sync_with_stdio(false);
    // untie in/out streams
    cin.tie(nullptr);
    return 0;
}();

class Solution {
private:
    unordered_map<int, int> umap;
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        int size = nums.size();
        for (int i = 0; i < size; i++){
            auto r = umap.find(target - nums[i]);
            if (r != umap.end()){
                return vector<int>{i, r->second};
            }
            umap[nums[i]] = i;
        }
    }
};

Discussion

前面那一段，用来解决频繁输出到stdout中很费时间的问题。

后面，用unordered_map实现快速查找，很神奇。

另外，把unordered_map<int, int> umap;的初始化放到函数外面，进一步提速（很trick）。