题目描述
Given an array of integers nums and an integer target, return indices of the two numbers such that they add up to target.
You may assume that each input would have exactly one solution, and you may not use the same element twice.
You can return the answer in any order.
Example1:
Input: nums = [2,7,11,15], target = 9 Output: [0,1] Output: Because nums[0] + nums[1] == 9, we return [0, 1].
Example2:
Input: nums = [3,2,4], target = 6 Output: [1,2]
Example3:
Input: nums = [3,3], target = 6 Output: [0,1]
Constraints:
- 2 <= nums.length <= 10^4
- -10^9 <= nums[i] <= 10^9
- -10^9 <= target <= 10^9
- Only one valid answer exists.
解题思路
- 常规算法O(n^2)
- 使用一种Hashmap，其中用具体的nums[i]做键，下标索引做value
##
class Solution {
public:
vector<int> twoSum(vector<int>& nums, int target) {
unordered_map<int, int> mmap;
vector<int> res;
for(int i = 0; i < nums.size(); i++){
if(mmap.find(target - nums[i]) != mmap.end()){
res.push_back(i);
res.push_back(mmap[target - nums[i]]);
}else{
mmap[nums[i]] = i;
}
}
return res;
}
};
