LeetCode Solution-1-蒲公英云

LeetCode Solution-1

蔚落 2023-08-17 17:45 214阅读 0赞

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:
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。

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。
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：

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

先保存原数组为num，然后对原数组按从小到大进行排序；
设定2个值begin和end分别作为开始和结束值的下标；
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；
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:

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;
}