寻找和为定值的N个数

1.寻找和为定值的N个数

#include<iostream>#include<list>using namespace std;list<int> list1;void GetNSum(int *a,int k, int n){if(k == a[n-1]){list<int>::iterator it;for(it = list1.begin();it!=list1.end();it++){cout<<*it<<" ";}cout<<a[n-1]<<endl;return;}else if(n<=0) return;else {GetNSum(a,k,n-1);//---------------------------list1.push_front(a[n-1]);GetNSum(a,k-a[n-1],n-1);list1.pop_front();}}int main(){int a[10] = {-1,0,1,2,-1,-4};GetNSum(a,0,6);}

2.寻找和为定值的3个数（在寻找和为定值的n个数方法中对链表的长度进行限制）

#include<iostream>#include<list>using namespace std;list<int> list1;void GetNSum(int *a,int k, int n){if(k == a[n-1]){if(list1.size() == 2){list<int>::iterator it;for(it = list1.begin();it!=list1.end();it++){cout<<*it<<" ";}cout<<a[n-1]<<endl;return;}else return;}else if(n<=0) return;else {GetNSum(a,k,n-1);//---------------------------list1.push_front(a[n-1]);GetNSum(a,k-a[n-1],n-1);list1.pop_front();}}int main(){int a[10] = {-1,0,1,2,-1,-4};GetNSum(a,0,6);}

另一种方法，参考别的，自己也实现了下代码

思路：假设输入数组个数为n，每一个数用1bit表示，bit位为0表示没用使用这个数，bit为1 表示使用这个数，比如7位111，则使用了数组中a【0】、a【1】、a【2】中的三个数，首先找找 有三个1的，再看加和是否为指定的值（例子的加和默认为0）。

#include<iostream>
#include<vector>
using namespace std;
class Solution {
public:
int numof1(int i)
{
int result = 0;
while(i)
{
i = i&(i-1);
result++;
}
return result;
}
vector<vector<int> > threeSum(vector<int>& nums) {
vector<vector<int> >a;
int len = nums.size();
cout<<"nums.size()="<<nums.size()<<endl;
int bit = 1<<len;
cout<<"bit="<<bit<<endl;
for(int i = 1;i<bit;i++)
{
int sum = 0;
vector<int> tmp;
if(numof1(i)==3)
{
cout<<"i = "<<i<<" ";
for(int j=0;j<len;j++)
{
if((i&(1<<j)) != 0)
{
int tmpnum = i&(1<<j);
cout<<"tmpnum="<<tmpnum<<" ";
sum+=nums[j];
cout<<nums[j]<<"("<<j<<")"<<" ";
tmp.push_back(nums[j]);
}
}
cout<<"sum = "<<sum<<endl;
if(sum == 0)
{
a.push_back(tmp);
}
}
}
return a;
}
};
int main()
{
vector<int> test;
test.push_back(-1);
test.push_back(0);
test.push_back(1);
test.push_back(2);
test.push_back(-1);
test.push_back(-4);
class Solution A;
vector<vector<int> >a = A.threeSum(test);
cout<<a.size()<<endl;
for(int i = 0;i<a.size();i++)
{
cout<<a[i][1]<<" "<<a[i][2]<<" "<<a[i][3]<<endl;
}
}