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].
暴力解题:时间复杂度O(n^2)
class Solution {
public:
vector<int> twoSum(vector<int>& nums, int target) {
vector<int> vec;
for (int i = 0;i < nums.size();++i)
{
for (int j = 0;j < nums.size();++j)
{
if (nums[i] + nums[j] == target && i != j && j > i)
{
vec.push_back(i);
vec.push_back(j);
}
}
}
return vec;
}
};
看了大佬的解题思路,也理解了更为简单的解题方法:时间复杂度为O(n)
class Solution {
public:
vector<int> twoSum(vector<int>& nums, int target) {
unordered_map<int,int> mp; //要保存数组的值-下标,值可能是无序的,所以采用unordered_map
vector<int> vec;
for (int i = 0;i < nums.size(); ++i)
{
mp[nums[i]] = i; //用给定数据初始化map
}
for (int i = 0;i < nums.size(); ++i)
{
int t = target - nums[i]; //nums[i]为第一个数,t得到第二个数
if (mp.count(t) && mp[t] != i) //判断t是否符合要求
{
vec.push_back(i); //将所得两个数的下标存入目标容器
vec.push_back(mp[t]);
break; //如找到则退出搜索
}
}
return vec; //返回所得结果
}
};
更为简洁的方案:
class Solution {
public:
vector<int> twoSum(vector<int>& nums, int target) {
unordered_map<int,int> mp;
for (int i = 0;i < nums.size();++i)
{
if (mp.count(target - nums[i]))
{
return {mp[target - nums[i]], i}; //由于省去了单独的初始化过程,在这里要特别注意,这里是找到第二个数之后回去找到第一个数的结果,因此i代表第二个数的下标
}
mp[nums[i]] = i;
}
return {};
}
};