TwoSum

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].

解法1：暴力破解

int* twoSum(int* nums, int numsSize, int target) {
    int *index = (int*)malloc(sizeof(int)*2);
    for (int i = 0; i < numsSize - 1; i++)
    {
        for (int j = i + 1; j < numsSize; j++)
        {
            if (nums[i] + nums[j] == target)
            {
                index[0] = i;
                index[1] = j;
                return index;
            }
            
        }
    }
    return index;
}

时间复杂度：O(n2)，用了双重循环

解法2：单循环map

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        map<int, int> map;
        for (int i = 0; i < nums.size(); i++)
        {
            int complement = target - nums[i];
            if (map.count(complement) > 0)
            {
                return { map[complement],i };
            }
            map[nums[i]] = i;
        }
        return {};
    }
};

时间复杂度：O(n)，只有一个循环

解法3：双循环map

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        map<int, int> map;
        for (int i = 0; i < nums.size(); i++)
        {
            map[nums[i]] = i;
        }

        for (int i = 0; i < nums.size(); i++)
        {
            int complement = target - nums[i];
            if (map.count(complement) > 0 && map[complement] != i)
            {
                return { i,map[complement] };
            } 
        }
        return {};
    }
};

时间复杂度：O(n)，只有一个循环