1. Two Sum

Problem:

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].

思路1
利用哈希表。

  1. 首先新建一个哈希表(unordered_map),存储输入数组元素及对应的下标。
  2. 对数组元素进行遍历,对每个值得到其和目标值的差值(dif),在数组中寻找dif,如果存在且下标不重复即得到结果,若遍历后无结果则返回空数组。

Solution (C++):

vector<int> twoSum(vector<int>& nums, int target) {
    unordered_map<int, int> num_id;
    for (int i = 0; i < nums.size(); i++) {
        num_id[nums[i]] = i;
    }

    for (int i = 0; i < nums.size(); i++) {
        int dif = target - nums[i];
        auto dif_id = num_id.find(dif);
        if (dif_id != num_id.end() && dif_id->second != i) 
            return vector<int>{i, dif_id->second};
    }
    return vector<int>();
}

性能

Runtime: 8 ms  Memory Usage: 10.5 MB

思路2

对数组排序

  1. 先保存原数组为num,然后对原数组按从小到大进行排序;
  2. 设定2个值begin和end分别作为开始和结束值的下标;
  3. 比较nums[beign]和nums[end]的和sum与target的大小,若sum < target,则说明2数偏小,begin++;若sum > target,则说明两数之和偏大,end--;如果sum = target,则进入4;
  4. 在num中分别查找nums[begin]和nums[end]对应的下标值,并判断2数是否相等,若不相等即是最后的输出。

Solution (C++):

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

性能

Runtime: 8 ms  Memory Usage: 9.5 MB

posted @ 2019-09-06 17:23  littledy  阅读(160)  评论(0编辑  收藏  举报