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更新时间：2022-10-03 13:44:17
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.
Example:
Given nums = [2, 7, 11, 15], target = 9, Because nums[0] + nums[1] = 2 + 7 = 9, return [0, 1].
题目大意：
在一个数组中找出2个元素的和等于目标数，输出这两个元素的下标。
思路：
最笨的办法喽，双循环来处理。时间复杂度O(n*n)。
代码如下：
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class Solution {
public:
vector<int> twoSum(vector<int>& nums, int target) {
vector<int> result;
int i,j;
for(i = 0; i < nums.size();i++)
{
for(j = i+1; j < nums.size();j++)
{
if(nums[i] + nums[j] == target)
{
result.push_back(i);
result.push_back(j);
break;
}
}
}
return result;
}
}; |
参考他人的做法：https://discuss.leetcode.com/topic/3294/accepted-c-o-n-solution
采用map的键值，把元素做键，把元素的下标做值。
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vector<int> twoSum(vector<int> &numbers, int target)
{ //Key is the number and value is its index in the vector.
unordered_map<int, int> hash;
vector<int> result;
for (int i = 0; i < numbers.size(); i++) {
int numberToFind = target - numbers[i];
//if numberToFind is found in map, return them
if (hash.find(numberToFind) != hash.end()) {
result.push_back(hash[numberToFind]);
result.push_back(i);
return result;
}
//number was not found. Put it in the map.
hash[numbers[i]] = i;
}
return result;
} |