1048. Find Coins (25) -- 二分法，寻找和为定值的两个数 (Two Sum - leetcode)

1048. Find Coins (25)

时间限制

50 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

Eva loves to collect coins from all over the universe, including some other planets like Mars. One day she visited a universal shopping mall which could accept all kinds of coins as payments. However, there was a special requirement of the payment: for each bill, she could only use exactly two coins to pay the exact amount. Since she has as many as 10⁵ coins with her, she definitely needs your help. You are supposed to tell her, for any given amount of money, whether or not she can find two coins to pay for it.

Input Specification:

Each input file contains one test case. For each case, the first line contains 2 positive numbers: N (<=10⁵, the total number of coins) and M(<=10³, the amount of money Eva has to pay). The second line contains N face values of the coins, which are all positive numbers no more than 500. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print in one line the two face values V₁ and V₂ (separated by a space) such that V₁ + V₂ = M and V₁ <= V₂. If such a solution is not unique, output the one with the smallest V₁. If there is no solution, output "No Solution" instead.

Sample Input 1:

8 15
1 2 8 7 2 4 11 15

Sample Output 1:

4 11

Sample Input 2:

7 14
1 8 7 2 4 11 15

Sample Output 2:

No Solution

---

提交代码  http://www.patest.cn/contests/pat-a-practise/1048

// http://www.patest.cn/contests/pat-a-practise/1048
// 1048. Find Coins (25)
#include <cstdio>  
#include <iostream> 
#include <vector>  
#include <set>
#include <algorithm>
using namespace std;

#define N  100005

int n , m ;
int a[N] ;

bool bsearch(int left,int right , int val)
{
	while(left <= right)
	{
		int mid = (left+right) / 2 ;
		if(a[mid] == val)
		{
			return true;
		}
		else if(a[mid] < val)
		{
			left = mid +1 ;
		}else{
			right = mid - 1;
		}
	}
	return false;
}
int main()
{
	//freopen("in.txt", "r", stdin);
	while( scanf("%d%d" , &n , &m) != EOF)
	{
		int i ,j ;
		for(i = 0 ; i < n ; i ++)
		{
			scanf("%d",&a[i]);
		}
		sort(a,a+n);
		bool flag = false;
		int val = -1 ;
		for(i = 0 ;i < n-1;i++)
		{
			if(a[i] <m)
			{
				int tmp = m - a[i];
				if(bsearch(i+1,n-1, tmp))
				{
					flag = true;
					val = a[i];
					break;
				}
			}else{
				break;
			}
			if(flag == true)
				break;
		}
		if(flag)
			printf("%d %d\n",val , m - val);
		else
			printf("No Solution\n");
	}
	return 0;
}
/*
Sample Input 1:
8 15
1 2 8 7 2 4 11 15
Sample Output 1:
4 11
Sample Input 2:
7 14
1 8 7 2 4 11 15
Sample Output 2:
No Solution
*/

解法2：

利用排序，二分查找

其中一个map<int, int>用来记录某个数字个数，主要是考虑v1 == v2的特殊情况

另外用到了map<int, int>主要是避免重复v2，因为原序列数字是可以重复的，否则易出现超时现象

多次用到了 nlogn 的时间复杂度（包括排序，二分查找），总的时间复杂度是nlogn , 不过 空间复杂度比较大

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <map>
using namespace std;

const int N = 100001;
int n , m ;

int coins[N];
int sub_coins[N];
map<int, int> mp;

int m_bsearch(int a[], int num, int left, int right)
{
  if(left > right)
    return -1;
  while(left <= right)
  {
    int mid = (left + right) / 2;
    if(a[mid] == num)
      return mid;
    else if(a[mid] < num) {
      left = mid + 1;
    } else {
      right = mid - 1;
    }
  }
  return -1 ;
}

bool cmp(int a, int b)
{
  return a > b;
}

int main()
{
  //freopen("in.txt", "r", stdin);
  scanf("%d%d", &n , &m);
  int i;
  for(i = 0; i < n; i++)
  {
    scanf("%d", &coins[i]);
    mp[coins[i]] ++;
    sub_coins[i] = m - coins[i];
  }
  sort(coins, coins + n);
  sort(sub_coins, sub_coins + n, cmp);

  map<int, int> mp_visit;
  for(i = 0; i < n ; i++)
  {
    int v1 = m - sub_coins[i];
    int v2 = sub_coins[i];
    if(mp_visit[v2] == 1)
      continue;
    if( m_bsearch(coins, sub_coins[i], 0, n - 1) != -1)
    {
      
      if( (v1 < v2) || (v1 == v2 && mp[v1] >= 2) )
      {
        printf("%d %d\n", v1, v2);
        return 0;
      }
    }
    mp_visit[v2] = 1;
  }
  printf("No Solution\n");
  return 0;
}

方法3：

采用前后指针法

总的空间复杂度是O(1), 时间复杂度是 O(nlogn + n) = O(nlogn)

#include <cstdio>
#include <iostream>
#include <algorithm>

using namespace std;

const int N = 100001;
int n , m ;
int coins[N];

int main()
{
  //freopen("in.txt", "r", stdin);
  scanf("%d%d", &n , &m);
  int i;
  for(i = 0; i < n; i++)
    scanf("%d", &coins[i]);
  sort(coins, coins + n); // 排序
  int begin = 0, end = n - 1;
  while( begin < end)
  {
    int v1 = coins[begin];
    int v2 = coins[end];
    if(v1 + v2 == m) {
      printf("%d %d\n", v1, v2);
      return 0;
    } else if(v1 + v2 < m) {
      begin ++;
    } else {
      end --;
    }
  }
  printf("No Solution\n");
  return 0;
}

--------

解法4： 类似题Two Sum - leetcode)，使用map结构处理结果

#include <map>
#include <cstdio> 
#include <cstdlib>
#include <algorithm>  
using namespace std;

#define N  100005  

int n, m;
int a[N];

int main()
{
  // freopen("in.txt", "r", stdin);  
  while (scanf("%d%d", &n, &m) != EOF)
  {
    for (int i = 0; i < n; i++)
    {
      scanf("%d", &a[i]);
    }
    sort(a, a + n);
    bool flag = false;
    map<int, int> mp;
    int v1, v2;
    for (int i = 0; i < n; i++)
    {
      if (mp[m - a[i]] == 0)
      {
        mp[a[i]] = 1;
      }
      else{
        if (m - a[i] <= a[i])
        {
          if (flag == false)
          {
            v1 = m - a[i];
            v2 = a[i];
            flag = true;
          }
          else{
            if (m - a[i] < v1){
              v1 = m - a[i];
              v2 = a[i];
            }
          }
        }
      }
    }

    if (flag)
      printf("%d %d\n", v1, v2);
    else
      printf("No Solution\n");
  }
  return 0;
}