leetcode1. 两数之和、15. 三数之和、18. 四数之和

1. 两数之和

注意，这个要返回的是对应数字的下标。

 

2022/09的C++代码

class Solution {
public:vector<int> twoSum(vector<int>& nums, int target) {int size=nums.size();unordered_map<int,int> mapmap;for(int i=0;i<size;i++){auto it=mapmap.find(target-nums[i]);if(it == mapmap.end()){mapmap[nums[i]]=i;//mapmap.insert(pair<int, int>(nums[i], i));//另一种添加方法
            }else{return {it->second,i};}}  return {};}
};

2025/10的Java代码

class Solution {public int[] twoSum(int[] nums, int target) {int n = nums.length;int[] res = new int[2];HashMap<Integer,Integer> record = new HashMap<>();for(int i = 0;i < n;++i){if(record.containsKey(nums[i]) && nums[i] * 2 == target){res[0] = record.get(nums[i]);res[1] = i;return res;}int minus = target - nums[i];if(record.containsKey(minus) && i != record.get(minus)){//保证不是同一个元素用两次res[0] = i;res[1] = record.get(minus);return res;}record.put(nums[i],i);}return res;}
}

 

 

 

15. 三数之和

注意，这个返回的是具体数字，不能重复，比如[ -1 , 0 , 1 ] 和 [ -1 , 1 , 0 ]就是重复的。

以下为无法通过的超时解，每次添加前都查找是否已存在太耗时间了。

class Solution {public List<List<Integer>> threeSum(int[] nums) {List<List<Integer>> res = new ArrayList<>();int n = nums.length;Arrays.sort(nums);for(int i = 0;i < n;++i){if(nums[i] > 0)  break;if(i != 0 && nums[i] == nums[i-1])  continue;HashMap<Integer,Integer> record = new HashMap<>();//记录本轮目标值为(0-nums[i])的哈希表int target1 = 0 - nums[i];for(int j = i + 1;j < n;++j){if(record.containsKey(nums[j]) && nums[j] * 2 == target1){List<Integer> thisRes = new ArrayList<>(Arrays.asList(nums[i],nums[j],nums[j]));if(!res.contains(thisRes))  res.add(thisRes);continue;}int target2 = target1 - nums[j];if(record.containsKey(target2) && j != record.get(target2)){List<Integer> thisRes = new ArrayList<>(Arrays.asList(nums[i],nums[record.get(target2)],nums[j]));//res.add(new ArrayList<>(Arrays.asList(nums[i],nums[record.get(target2)],nums[j])));if(!res.contains(thisRes))  res.add(thisRes);continue;}record.put(nums[j],j);}}return res;}
}

以上代码改成HashSet，最后再转为List返回就可以通过。保证添加到HashSet时ArrayList是有序的。

但还是太低效。用时417ms，击败5.01%，内存52.50MB，击败5.00%。

class Solution {public List<List<Integer>> threeSum(int[] nums) {HashSet<List<Integer>> resSet = new HashSet<>();int n = nums.length;Arrays.sort(nums);for(int i = 0;i < n;++i){if(nums[i] > 0)  break;if(i != 0 && nums[i] == nums[i-1])  continue;HashMap<Integer,Integer> record = new HashMap<>();//记录本轮目标值为(0-nums[i])的哈希表int target1 = 0 - nums[i];for(int j = i + 1;j < n;++j){if(record.containsKey(nums[j]) && nums[j] * 2 == target1){resSet.add(new ArrayList<>(Arrays.asList(nums[i],nums[j],nums[j])));continue;}int target2 = target1 - nums[j];if(record.containsKey(target2) && j != record.get(target2)){resSet.add(new ArrayList<>(Arrays.asList(nums[i],nums[record.get(target2)],nums[j])));continue;}record.put(nums[j],j);}}return new ArrayList<>(resSet);}
}