[LeetCode]1. Two Sum

1. 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.

Example:

Given nums = [2, 7, 11, 15], target = 9,Because nums[0] + nums[1] = 2 + 7 = 9,return [0, 1].

分析

1. 最直接的暴力解决，但速度慢
2. 使用哈希表，用于记录每个数对应的下标，复杂度O(n)
3. 如果返回的不是下标，则还可以先排序然后使用两边夹逼方式寻找；但本题要寻找下标，所以这个方法行不通

源码

暴力解决

    vector<int> twoSum(vector<int>& nums, int target) {        vector<int> ret;        for(vector<int>::const_iterator it = nums.begin(); it < nums.end(); it++) {            for(vector<int>::const_iterator inner = it + 1; inner < nums.end(); inner++) {                if (target == (*it + *inner)) {                    ret.push_back(it - nums.begin());                    ret.push_back(inner - nums.begin());                    break;                }            }        }        return ret;    }

使用哈希表

vector<int> twoSum(vector<int>& nums, int target) {        vector<int> ret;        unordered_map<int,int> mapper;        for(int i = 0; i < nums.size(); i++) { // 遍历原数组            mapper[nums[i]] = i;            // [0,4,3,3] target = 6             int gap = target - nums[i];            // mapper[gap] > i 肯定是要往比索引i大的方向查找 比如[3,2,4] target = 6            // 如果不加入这个条件限制，就会返回[0,0],因为6-3=3，而3在哈希表里，不能是自己了            if(mapper.find(gap) != mapper.end() && mapper[gap] > i) {                 ret.push_back(i);                ret.push_back(mapper[gap]);                break;            }        }        return ret;    }