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.
Accepted
3.6M
Submissions
7.8M

class Solution {
    public int[] twoSum(int[] nums, int target) {
        int[] indices = new int[2];
        HashMap<Integer, Integer> tmpmap = new HashMap<Integer, Integer>(); //  key为nums的值，value为nums下标

        for (int i = 0; i < nums.length; ++i){
            if(tmpmap.containsKey(Integer.valueOf(nums[i]))){
                indices[0] = tmpmap.get(Integer.valueOf(nums[i])).intValue();
                indices[1] = i;
                break;
            }
            tmpmap.put(Integer.valueOf(target-nums[i]), Integer.valueOf(i));
        }

        return indices;
    }
}

class Solution {
public:
    /*vector<int> twoSum(vector<int>& nums, int target) {
        vector<int> indices(2);
        int len = nums.size();
        for(int i = 0; i < len; ++i){
            for(int j = i+1; j < len; ++j){
                if(nums[i] + nums[j] == target){
                    indices[0] = i;
                    indices[1] = j;
                    break;
                }
            }
        }
        return indices;
    }*/
    vector<int> twoSum(vector<int>& nums, int target){
    vector<int> twoSum;
    map<int, int> tmpmap;//键值为nums的值，变量值为nums下标
    
    for(int i=0;i<nums.size();++i){
        if(tmpmap.count(nums[i])!=0){
            twoSum.push_back(tmpmap[nums[i]]);
            twoSum.push_back(i);
            break;
         }
        tmpmap[target-nums[i]]=i;
    }
    return twoSum;
    }
};