leetcode刷题记录
第一题 Two Sum
题目描述: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, and you may not use the same element twice.
示例:Given nums = [2, 7, 11, 15], target = 9,
Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].
解答:
初始想法是枚举法,对两个集合进行遍历,最终可以通过,但效率很低,只超过13%的人。这是万万不行的。
初始代码:
class Solution {
public:
vector<int> twoSum(vector<int>& nums, int target) {
vector<int> v;
for(int i=0;i<nums.size()-1;++i)
{
for(int j=i+1;j<nums.size();++j)
{
if(nums[i]+nums[j]==target)
{
v.push_back(i);
v.push_back(j);
}
}
}
return v;
}
};
改进
使用HashTable implementation,可以将时间代价控制到O(n);
static int lambda_0 = []() { std::ios::sync_with_stdio(false); cin.tie(NULL); return 0; }();
class Solution {
public:
vector<int> twoSum(vector<int>& nums, int target) {
vector<int> indices;
map<int,int> valueIndices;
for(int i =0; i < nums.size(); i++){
if(valueIndices.find(target-nums[i]) != valueIndices.end()){
indices.push_back(i);
indices.push_back(valueIndices[target-nums[i]]);
return indices;
}
valueIndices[nums[i]] = i;
}
return indices;
}
};
时间复杂度减少到O(n),