题目

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.

Example 1:

Input: nums = [2,7,11,15], target = 9
Output: [0,1]
Output: Because nums[0] + nums[1] == 9, we return [0, 1].
Example 2:

Input: nums = [3,2,4], target = 6
Output: [1,2]
Example 3:

Input: nums = [3,3], target = 6
Output: [0,1]

Constraints:

2 <= nums.length <= 103
-109 <= nums[i] <= 109
-109 <= target <= 109
Only one valid answer exists.

解法：哈希表

遍历数组，对于每个数nums[i]：

• 如果target-nums[i]在哈希表中，直接返回答案
• 否则存入哈希表

时间复杂度O(n)，空间复杂度O(n)

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        unordered_map<int, int> h;
        for (int i = 0; i < nums.size(); i++) {
            int r = target - nums[i];
            if (h.count(r)) return {h[r], i};
            h[nums[i]] = i;
        }
        return {};
    }
};