LeetCode 15. 3Sum; 16. 3Sum Closest; 259. 3Sum Smaller; 18. 4Sum

15. 3Sum

Two Sum 的 follow up

Two Sum 使用hashtable做到O(n)时间复杂度

所以看到这道题，第一想法是固定一个元素，剩下的用 Two Sum 处理。但是由于这道题有重复元素存在，最后去重会TLE，因此不能这样做。

在一个有序数组里寻找加和为给定值的两个元素，我们可以使用 Two Pointers 从两侧向中间来寻找。

对于这道题，我们可以先对数组进行排序，固定一个数nums[i]，用 Two Pointers 寻找加和等于 -nums[i] 的两个数。每次两个pointer都跳过重复的元素，固定的nums[i]也跳过重复的元素，这样最后答案里所有的解都是唯一的了。

class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        vector<vector<int>> res;
        sort(nums.begin(),nums.end());
        for (int i=0;i<nums.size();++i){
            int target= -nums[i];
            int low=i+1, high=nums.size()-1;
            while (low<high){
                int sum=nums[low]+nums[high];
                if (sum<target) ++low;
                else if (sum>target) --high;
                else{ 
                    vector<int> tri({nums[i],nums[low],nums[high]});
                    res.push_back(tri);
                    
                    //skip the same value as nums[low]
                    while (low<high && nums[low]==tri[1]) ++low;
                    
                    //skip the same value as nums[high]
                    while (low<high && nums[high]==tri[2]) --high;
                }
            }
            //skip the same value as nums[i]
            while (i+1<nums.size() && nums[i+1]==nums[i])
                ++i;
        }
        return res;
    }
};

 

16. 3Sum Closest

和 3 Sum 几乎一模一样，复习一下。

class Solution {
public:
    int threeSumClosest(vector<int>& nums, int target) {
        if (nums.size()<3) return 0;
        int res=nums[0]+nums[1]+nums[2];
        sort(nums.begin(),nums.end());
        
        for (int i=0;i<nums.size();++i){
            if (i>=1 && nums[i]==nums[i-1]) continue;
            int low=i+1, high=nums.size()-1;
            
            while (low<high){
                int curSum=nums[i]+nums[low]+nums[high];
                if (curSum==target) return curSum;
                if (abs(curSum-target)<abs(res-target))
                    res = curSum;
                if (curSum<target) ++low;
                else if (curSum>target) --high;
            }
        }
        return res;
    }
};

 

259. 3Sum Smaller

curSum<target 时，low不变，high从high到low+1都是可行的，所以+=high-low。

class Solution {
public:
    int threeSumSmaller(vector<int>& nums, int target) {
        int cnt=0;
        sort(nums.begin(),nums.end());
        
        for (int i=0;i<nums.size();++i){
            int low=i+1, high=nums.size()-1;
            while (low<high){
                int curSum=nums[i]+nums[low]+nums[high];
                if (curSum<target){
                    cnt += high-low; // the third num can be any element in (low,high]
                    ++low;
                }else --high;
            }
        }
        return cnt;
    }
};

 

18. 4Sum

还是一样的套路，把之前的写法跟精简了一下：

class Solution {
public:
    vector<vector<int>> fourSum(vector<int>& nums, int target) {
        vector<vector<int>> res;
        sort(nums.begin(),nums.end());
        for (int i=0;i<nums.size();++i){
            if (i>0 && nums[i]==nums[i-1]) continue;
            for (int j=i+1;j<nums.size();++j){
                if (j>i+1 && nums[j]==nums[j-1]) continue;
                int low=j+1, high=nums.size()-1;
                while (low<high){
                    int sum=nums[i]+nums[j]+nums[low]+nums[high];
                    if (sum==target){
                        res.push_back({nums[i],nums[j],nums[low],nums[high]});
                        ++low; --high;
                        while (low<high && nums[low]==nums[low-1]) ++low;
                        while (low<high && nums[high]==nums[high+1]) --high;
                    }else if (sum<target) ++low;
                    else --high;
                }
            }
        }
        return res;
    }
};