LeetCode 1. Two Sum Java

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本文介绍了一种解决两数之和问题的有效算法。给定一个整数数组和一个目标值，通过一次遍历结合哈希表的方式找到数组中两个数相加等于目标值的下标，并确保不重复使用同一元素。提供了两种实现方式，一种是基于Java的解决方案，另一种是基于C++的实现。

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;
    }
};