【Leetcode】3Sum,3sum closest,3sum smaller, 4Sum

3Sum

// Author : yqtao// Date   : 2016.6.17// Email  : [e-mail]/***************************************************************************************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:* * Elements in a triplet (a,b,c) must be in non-descending order. (ie, a ≤ b ≤ c)* 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)**********************************************************************************************/

class Solution {public:    vector<vector<int>> threeSum(vector<int>& nums) {        vector<vector<int>> res;        if (nums.size()<=2) return res;        sort(nums.begin(),nums.end());        for (int i=0;i<nums.size()-2;i++) {            if (i>0&&nums[i]==nums[i-1])  continue;            int a=nums[i],low=i+1,high=nums.size()-1;            while (low<high) {                int b=nums[low],c=nums[high];                if (a+b+c==0) {                    vector<int> v;                    v.push_back(a);                    v.push_back(b);                    v.push_back(c);                    res.push_back(v);                }                if (a+b+c>0) {                    while (low<high&&nums[high]==nums[high-1]) high--;                    high--;                }                else {                    while (low<high&&nums[low]==nums[low+1]) low++;                    low++;                }            }        }        return res;    }};

3Sum closest

/************************************************************************************ 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).************************************************************************************/

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

3Sum smaller

Given an array of n integers nums and a target, find the number of index triplets i, j, k with 0 <= i < j < k < n that satisfy the condition nums[i] + nums[j] + nums[k] < target.For example, given nums = [-2, 0, 1, 3], and target = 2.Return 2. Because there are two triplets which sums are less than 2:[-2, 0, 1][-2, 0, 3]

解析：因为要找三个数相加比target小的,因此只需要移动最右边的指针即可。

int threeSumSmaller(vector<int>&nums, int target){      int size = nums.size();      sort(nums.begin(),nums.end());      int cnt = 0;      for(int i = 0;i < size-2;i++){          int j = i+1,k = size - 1;          while(j < k){             int sum = nums[i] + nums[j] + nums[k];                        if(sum < target)cnt++;             k--;    //仅仅移动此指针        }               }      return cnt;  }  

4 Sum

/************************************************************************************ 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:** Elements in a quadruplet (a,b,c,d) must be in non-descending order. (ie, a ≤ b ≤ c ≤ d)* 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)************************************************************************************/

解析：可以将4sum的问题变成3sum的问题进行求解。

#include "stdafx.h"#include <iostream>#include <vector>#include <algorithm>using namespace std;vector<vector<int>> threeSum(vector<int>&num, int target);vector<vector<int>> fourSum(vector<int>&num, int target){    vector<vector<int>> result;    if (num.size() < 4) return result;    sort(num.begin(), num.end());//首先进行排序    for (int i = 0; i < num.size() - 3; i++)    {        if (i > 0 && num[i - 1] == num[i])            continue;        //复制到n中，然后用3sum方法解决        vector<int>n(num.begin() + i + 1, num.end());        vector<vector<int>> ret = threeSum(n, target - num[i]);        for (int j = 0; j < ret.size(); j++)        {            ret[j].insert(ret[j].begin(), num[i]);//在结果前插入num[i]            result.push_back(ret[j]);        }    }    return result;}vector<vector<int>> threeSum(vector<int>&num,int target){    vector<vector<int>> result;//返回二维数组    sort(num.begin(), num.end());//首先进行排序    int n = num.size();    for (int i = 0; i < n - 2; i++)// 因为三个数，所以第一个数最大只能为len-3    {        //去掉重复值,这里一定要注意，i如果与前一个i所在值相同，则        //不必计算了，直接另i++        //不能写成num[i]==num[i+1]        if (i > 0 && num[i-1] == num[i])            continue;        int a = num[i];        int low = i + 1;        int high = n - 1;        while (low < high)        {            int b = num[low];            int c = num[high];            if (a+b+c==target)//满足条件，将其加入数组中            {                vector<int>v;                v.push_back(a);                v.push_back(b);                v.push_back(c);                result.push_back(v);                //继续寻找下一个满足条件的值，去重值                while (low < high&&num[low] == num[low + 1]) low++;                low++;                while (high > low&&num[high] == num[high - 1]) high--;                high--;            }            else if (a + b + c>target)            {                while (high > low&&num[high] == num[high - 1])                    high--;                high--;            }            else            {                while (low < high&&num[low] == num[low + 1])                     low++;                low++;            }        }    }    return result;}//打印二维数组void printMatrix(vector<vector<int>> &matrix){    //基于范围for循环，等效于下面的for    for (auto c : matrix)    {        cout << "[";        for (auto r : c)            cout << r << " ";        cout <<"]"<< endl;    }    /*for (int i = 0; i < matrix.size(); i++)    {    cout << "[";    for (int j = 0; j < matrix[i].size(); j++)    cout << matrix[i][j] << "  ";    cout << "]" << endl;    }*/}int main(){    vector<int>num = { 1, 0, -1, 0, - 2, 2 };    vector<vector<int>>result = fourSum(num,0);    printMatrix(result);    return 0;}