【LeetCode 1_数组_哈希表】Two Sum

解法一：O(N)

vector<int> twoSum(vector<int>& nums, int target)
{
    unordered_map<int, int> numsHash;
    vector<int> res;

    for (int i = 0; i < nums.size(); ++i) {
        int num_last = target - nums[i];
        if (numsHash.find(num_last) != numsHash.end()) {
            res.push_back(numsHash[num_last] + 1);
            res.push_back(i + 1);
            break;
        } else {
            numsHash[nums[i]] = i;
        }
    }
    return res;
}

解法二：O(NlogN)

vector<int> twoSum(vector<int>& nums, int target)
{
    vector<int> res;
    int lpos = 0, rpos = nums.size() - 1;
    sort(nums.begin(), nums.end());
    while (lpos < rpos) {
        int num = nums[lpos] + nums[rpos];
        if (num == target) {
            res.push_back(lpos + 1);
            res.push_back(rpos + 1);
        } else if (num > target) {
            rpos--;
        } else {
            lpos++;
        }
    }
    return res;
}

解法二也是不错的思路，空间复杂度O(1)，但这个解法会超时。