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].
题目大意:给一串整形数组和一个整数,返回两个数下标,它们的和为给定的整数。保证只有一种,并且不能使用同一个数两次。
①暴力求解
class Solution {
public:
vector<int> twoSum(vector<int>& nums, int target) {
vector<int> result;
for(int i = 0; i < nums.size(); i++){
for(int j = i+1; j < nums.size(); j++){
if((nums[i] + nums[j]) == target){
result.push_back(i);
result.push_back(j);
return result;
}
}
}
return result;
}
};
- 时间复杂度为O(n^2)
②用哈希表存,用空间换时间,只遍历一个数,剩下的把target减去这个得到,再判断它在Hash表里是否存在。时间复杂度为O(n)
class Solution {
public:
vector<int> twoSum(vector<int>& nums, int target) {
unordered_map<int, int> hash;
vector<int> result;
for(int i = 0; i < nums.size(); i++){
if(hash.find(target-nums[i]) != hash.end()){
result.push_back(i);
result.push_back(hash[target-nums[i]]);
break;
}
hash[nums[i]] = i;
}
return result;
}
};
class Solution {
public:
vector<int> twoSum(vector<int>& nums, int target) {
unordered_map<int, int> hash;
for(int i = 0; i < nums.size(); i++){
if(hash.find(target-nums[i]) != hash.end()){
return {i, hash[target-nums[i]]};
}
hash[nums[i]] = i;
}
return {};
}
};