标签:none 存在 integer put 常用 one for 否则 targe
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]
这种是最容易想到的思路,比较暴躁,复杂度$O(N^2)$。
题目对数a、b求和,但是返回的是等于target时a、b的下标,于是想到将数组下标与对应值作一个映射表。C++使用unordered_map
关联容器可以实现键值与真值的映射。python中使用字典来实现类似功能。
template <class T, //键值类型
class T, // 映射类型
class hash = hash<key>, //哈希函数对象类型
class Pred = equal_to <key>, //相等比较函数对象类型
class Alloc = allocator < pair<cosnt key, T> > //alloctor类
>
count
size_type count (const key_type& k) const;
unordered_map
的数据以pair<const Key, T>保存,first是键值(key value),second是映射值(the mapped value)。赋值语句m[key value] = the mapped value
。
vector<int> twoSum(vector<int>& nums, int target) {
vector<int> results;
for(int i=0;i<nums.size();i++)
{
for(int j=i+1;j<nums.size();j++)
{
if(nums[i]+nums[j]==target)
{
results.push_back(i);
results.push_back(j);
return results;
}
else
{
continue;
}
}
}
}
vector<int> twoSum(vector<int>& nums, int target) {
unordered_map<int,int> m;
vector<int> results;
//数组中的值作为map的键值,其下标作为对应的映射值
for(int i=0;i<nums.size();i++)
{
m[nums[i]] =i;
}
for(int i = 0;i<nums.size();i++)
{
int t = target - nums[i];
if(m.count(t) && m[t] != i) // 不能使用同样的数两次
{
results.push_back(i);
results.push_back(m[t]);
break;
}
}
return results;
}
def twoSum(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: List[int]
"""
#建立字典
table ={nums[i] : i for i in range(len(nums))}
results = []
for i in range(len(nums)):
t = target - nums[i]
if table.get(t) is not None and (table.get(t) != i):
results = [i, table.get(t)]
break;
return results
标签:none 存在 integer put 常用 one for 否则 targe
原文地址:https://www.cnblogs.com/Jessey-Ge/p/10847901.html