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[
4000
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;
}
};