LeetCode hot100-15 三数之和

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class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        vector<vector<int>> n;
        sort(nums.begin(), nums.end());
        int size = nums.size();
        if (size < 3 || nums[0] > 0 || nums[size-1] < 0) return n;
        for(int i = 0 ;i < size - 2 && nums[i] <= 0;i++)
        {
            if(i > 0 && nums[i] == nums [i-1]) continue;
            int left = i+1;
            int right = size-1;
            while(left < right)
            {
                int sum =nums[i] + nums[left] + nums[right];
                if(sum == 0)
                {
                    n.push_back({nums[i],nums[left],nums[right]});
                    while( left < right && nums[left] == nums[left+1]) left++;
                    while( left < right && nums[right] == nums[right-1]) right--;
                    left++;
                    right--;
                }
                else if(sum < 0)
                {
                    left++;
                }
                else{
                    right--;
                }
            }
        }
        return n;
    }
};

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class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        set<vector<int>> results;  // 用 set 去重
        
        for (int i = 0; i < nums.size() - 2; i++) {
            // 跳过重复的 i
            if (i > 0 && nums[i] == nums[i-1]) continue;
            
            unordered_set<int> seen;
            for (int j = i + 1; j < nums.size(); j++) {
                int complement = -nums[i] - nums[j];
                if (seen.count(complement)) {
                    results.insert({nums[i], complement, nums[j]});
                } else {
                    seen.insert(nums[j]);
                }
            }
        }
        
        // 将 set 转换为 vector
        return vector<vector<int>>(results.begin(), results.end());
    }
};

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// 方法1：用 count（更简洁）
if (seen.count(complement)) {
    // 存在
}

// 方法2：用 find（更明确）
if (seen.find(complement) != seen.end()) {
    // 存在
}