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, 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].
Solution
1. brute force
class Solution {
public:
vector<int> twoSum(vector<int>& nums, int target) {
int n = nums.size();
vector<int> res;
for (int i = 0; i < n; i++)
{
for (int j = i+1; j < n; j++)
{
if(nums[i]+nums[j] == target) {
res.push_back(i);
res.push_back(j);
return res;
}
}
}
return res;
}
};
2. hashmap
下面要用到unordered_map
unordered_map<Key,T>::iterator it;
it->first; // same as (*it).first (the key value)
it->second; // same as (*it).second (the mapped value)
代码:
class Solution {
public:
vector<int> twoSum(vector<int>& nums, int target) {
unordered_map<int,int> hmap;
vector<int> res;
for ( int i = 0; i < nums.size(); i++)
{
int input = nums[i]; //避免多次查找元素,先储存下来
int key = target - input; //计算关键码
auto it = hmap.find(key); // auto 简化变量初始化,根据后面的数据类型推测;find 通过给定关键码查找元素,如果key存在,则find返回key对应的迭代器,如果key不存在,则find返回unordered_map::end
if ( it != hmap.end()) //如果key存在,说明已经找到符合条件的值,终止循环并返回
{
res.push_back(it->second); //保存哈希表中的mapped value
res.push_back(i); //保存当前下标
break;
}
//如果没有找到key,继续建立hash table
hmap[input] = i; //保存nums[i]关键码所对应的下标值
}
return res;
}
};
解题思路
这道题最简单能想到的就是遍历所有的可能数对,这样的话时间复杂度是O(n2)的。为了降低时间复杂度,我们可以考虑使用hashmap。
hashmap:
如果想知道是否有两个数的和等于target,可以先计算关键码key=target-nums[i]的值,然后在表中查找是否存在。如果存在直接返回下标。如果不存在,将{num[i], i} 保存到hash table中,以便下次查找。时间复杂度O(n)