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].
思路蠻簡單，將數(shù)組雙重遍歷就好
vector<int> twoSum(vector<int>& nums, int target) {
vector<int> v;
for (int i = 0; i < nums.size(); 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;
}
更優(yōu)解：建立HashMap，對(duì)于當(dāng)前值如果HashMap中有對(duì)應(yīng)解可以直接返回結(jié)果，否則將當(dāng)前數(shù)值加入到HashMap中殷费。
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()) {
//+1 because indices are NOT zero based
result.push_back(hash[numberToFind] + 1);
result.push_back(i + 1);
return result;
}
//number was not found. Put it in the map.
hash[numbers[i]] = i;
}
return result;
}
