LeetCode#15* 3Sum && LeetCode#16 3Sum Closest && LeetCode#18 4Sum

【3Sum】
Given an array S of n integers, are there elements a, b, c in S such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.

Note: The solution set must not contain duplicate triplets.

For example, given array S = [-1, 0, 1, 2, -1, -4],

A solution set is:
[
[-1, 0, 1],
[-1, -1, 2]
]

题意：给定n个整数的数组S，是否在数组S中有元素a，b，C，使得A + B + C =0？在数组中找出独一无二的三元素组，使得他们之和为0。
注意：
在三元素组（A，B，C）中，必须满足非递减排序。（即A≤B≤C）
该解决方案集中一定不能包含重复的三元素组。
思路：先对数组进行排序，然后枚举第一个数的位置，查找另外两个数的和等于-nums[i]的组合，由于数组排好序了，所以可以从两边往中间走，当结果大于0的时候后边往后退一步，否则前边进一步，时间复杂度O(n^2)，所以时间复杂度为O（n^2）。

class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        sort(nums.begin(),nums.end());             //sort 排序
        vector<vector<int>>ans;
        if(nums.size()<3)return ans;
        int i=0,l,r;
        while(i<nums.size()-2)
        {
            l=i+1;
            r=nums.size()-1;
            while(l<r)
            {
                if(nums[l]+nums[r]<-nums[i])l++;
                else if(nums[l]+nums[r]>-nums[i])r--;
                else {
                    vector<int>temp;
                    temp.push_back(nums[i]);
                    temp.push_back(nums[l]);
                    temp.push_back(nums[r]);
                    ans.push_back(temp);
                    l++;
                    while(l<r&&nums[l]==nums[l-1])
                    {
                        l++;
                    }
                }
            }
            i++;
            while(i<nums.size()-2&&nums[i]==nums[i-1])
            {
                i++;
            }
        }
        return ans;
    }
};

【3Sum Closet】
Given an array S of n integers, find three integers in S such that the sum is closest to a given number, target. Return the sum of the three integers. You may assume that each input would have exactly one solution.

For example, given array S = {-1 2 1 -4}, and target = 1.
The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).

题意：给一个整数数组，找到三个数的和与给定target的值距离最短的那个和。
思路：此题与上面那个题基本类似，甚至更简单一些，只需要比较和的结果即可，碰到和等于target的时候就直接返回吧！！！

class Solution {
public:
    int abs(int x)
    {
        return x>0?x:-x;
    }
    int threeSumClosest(vector<int>& nums, int target) {
        int i;
        int l,r;
        int min=INT_MAX/2;
        int len=nums.size();
        sort(nums.begin(),nums.end());
        for(i=0;i<len-2;i++)
        {
            l=i+1;
            r=len-1;
            while(l<r)
            {
                int sum=nums[l]+nums[r]+nums[i];
                if(abs(sum-target)<abs(min-target))min=sum;
                if(sum==target)return target;
                else if(sum>target)r--;
                else l++;
            }
        }
        return min;
    }
};

【4Sum】
Given an array S of n integers, are there elements a, b, c, and d in S such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target.

Note: The solution set must not contain duplicate quadruplets.

For example, given array S = [1, 0, -1, 0, -2, 2], and target = 0.

A solution set is:
[
[-1, 0, 0, 1],
[-2, -1, 1, 2],
[-2, 0, 0, 2]
]

题意：给一个数组，和一个target整数，求数组中能够使得和为target的所有组合a,b,c,d并满足a<=b<=c<=d。
思路：与上面的3Sum很相似，只不过多了一重循环。

class Solution {
public:
    vector<vector<int>> fourSum(vector<int>& nums, int target) {
        int len=nums.size();
        int i,j,l,r;
        sort(nums.begin(),nums.end());
        vector<vector<int>>ans;
        set<vector<int>>temp1;
        for(i=0;i<len-3;i++)
        {
            for(j=i+1;j<len-2;j++)
            {
                l=j+1;
                r=len-1;
                while(l<r)
                {
                    if(nums[i]+nums[j]+nums[l]+nums[r]==target)
                    {
                        temp1.insert({nums[i],nums[j],nums[l],nums[r]});
                        l++;             //注意 ：找到一组之后要l++或者r-- 否则会陷入死循环
                    }
                    else if(nums[i]+nums[j]+nums[l]+nums[r]>target)r--;
                    else l++;
                }
            }
        }
        //set去重
        auto it=temp1.begin();
        for(;it!=temp1.end();it++)
            ans.push_back(*it);
        return ans;
    }
};

LeetCode#15* 3Sum && LeetCode#16 3Sum Closest && LeetCode#18 4Sum

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