1.Two Sum(c++)(附6ms O(n) accepted 思路和代码)

问题描述：

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].

思路1：最简单的思路就是暴力求解，即循环遍历所有的可能情形，复杂度为o(n^2). 耗时未知，leetcode: Time Limit Exceeded

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        vector<int> result;
        int n = nums.capacity();
        for(int i=0;i<n;++i)
        {
            int i1 = nums.at(i);
            for(int j=i+1;j<n;++j)
            {
                if(i1+nums.at(j)==target)
                {
                    result.push_back(i);
                    result.push_back(j);
                    return result;
                }
            }
        }
        return result;
    }
};

思路2：使用hash map，查询等各种操作都是常数级别的时间复杂度（最好不要使用map，map实现大都是使用红黑树，查询的时间复杂度是O(logn)),复杂度为O(n),leetcode: AC,16ms

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        vector<int> result;
        unordered_map<int,int> hash;
        int n = nums.size();
        for(int i=0;i<n;i++)
        {
            if(hash.find(target-nums[i])!=hash.end())
            {
                result.push_back(hash[target-nums[i]]);
                result.push_back(i);
                return result;
            }
            hash[nums[i]] = i;
        }
        return result;
    }
};

思路3（参考论坛中4ms的c代码）：别出机杼，逆向思考，第一步找出数组中与目标值的差值（这一步可以筛选部分明显不可能得到目标值的数），第二步则是遍历数组找到哪个数是差值。复杂度：O(n). leetcode:AC,6ms,超过99.57%的结果！

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        vector<int> result;
        int n = nums.size();
        int max=0, min=0;
        for (int i = 0; i < n; i++)
        {
            if (nums[i] > max)
                max = nums[i];
            if (nums[i] < min)
                min = nums[i];
        }
        
        int* repeat_tag = new int[max - min + 1]();
        int diff;
        for (int i = 0; i < n; i++)
        {
            diff = target - nums[i];
            if (diff <= max && diff >= min)
            {
                repeat_tag[diff - min]= i;
            }
        }
        for (int i = 0; i < n; i++)
        {
            if (repeat_tag[nums[i] - min] != 0)
            {
                if (i != repeat_tag[nums[i] - min])
                {
                    result.push_back(i);
                    result.push_back(repeat_tag[nums[i] - min]);
                    return result;
                }
            }
        }    
        return result;
    }
};