Valid Sudoku
Determine if a Sudoku is valid, according to: Sudoku Puzzles - The Rules.
The Sudoku board could be partially filled, where empty cells are filled with the character ‘.‘.
A partially filled sudoku which is valid.
判断数独当前状态是否合法
class Solution
{
public:
bool isValidSudoku(vector<vector<char>>& board)
{
int col[10][10],row[10][10],box[10][10];
int i,j,x,a;
memset(col,0,sizeof(col));
memset(row,0,sizeof(row));
memset(box,0,sizeof(box));
for (i=0;i<9;i++)
for (j=0;j<9;j++)
if (board[i][j]!='.')
{
x=board[i][j]-'0';
a=i/3*3+j/3;
if (row[i][x]==1 || col[j][x]==1 || box[a][x]==1) return false;
else row[i][x]=col[j][x]=box[a][x]=1;
}
return true;
}
};
Sudoku Solver
Write a program to solve a Sudoku puzzle by filling the empty cells.
Empty cells are indicated by the character ‘.‘.
You may assume that there will be only one unique solution.
A sudoku puzzle...
...and its solution numbers marked in red.
DFS解数独，保证唯一解
class Solution
{
int col[10][10],row[10][10],box[10][10];
public:
void solveSudoku(vector<vector<char> >& board)
{
memset(col,0,sizeof(col));
memset(row,0,sizeof(row));
memset(box,0,sizeof(box));
for (int i=0;i<9;i++)
for (int j=0;j<9;j++)
if (board[i][j]!='.')
{
col[j][board[i][j]-'0']=1;
row[i][board[i][j]-'0']=1;
box[i/3*3+j/3][board[i][j]-'0']=1;
}
dfs(board,0);
}
bool dfs(vector<vector<char> > & board,int key)
{
int i,j;
if (key==81) return true;
int x=key/9;
int y=key%9;
if (board[x][y]!='.')
return dfs(board,key+1);
else
{
int a=x/3*3+y/3;
for (int i=1;i<=9;i++)
if (col[y][i]==0 && row[x][i]==0 && box[a][i]==0)
{
col[y][i]=row[x][i]=box[a][i]=1;
board[x][y]=i+'0';
if (dfs(board,key+1)) return true;
col[y][i]=row[x][i]=box[a][i]=0;
board[x][y]='.';
}
}
return false;
}
};
Count and Say
The count-and-say sequence is the sequence of integers beginning as follows:
1, 11, 21, 1211, 111221, ...
1 is read off as "one or
1"11.
11 is read off as "two or
1s"21.
21 is read off as "one, then
2one 1" or 1211.
Given an integer n, generate the nth sequence.
Note: The sequence of integers will be represented as a string.
模拟过程
class Solution
{
public:
string countAndSay(int n)
{
vector<int>mark[n+1];
int temp;
char ch;
mark[1].push_back(1);
string ans;
int i,sum,j;
for (i=2;i<=n;i++)
{
temp=mark[i-1][0];
sum=1;
for (j=1;j<mark[i-1].size();j++)
if (mark[i-1][j]!=temp)
{
mark[i].push_back(sum);
mark[i].push_back(temp);
temp=mark[i-1][j];
sum=1;
}
else sum++;
mark[i].push_back(sum);
mark[i].push_back(temp);
}
ans="";
for (i=0;i<mark[n].size();i++)
{
ch=mark[n][i]+'0';
ans=ans+ch;
}
return ans;
}
};
Combination Sum
Given a set of candidate numbers (C) and a target number (T), find all unique combinations in C where
the candidate numbers sums to T.
The same repeated number may be chosen from C unlimited number of times.
Note:
- All numbers (including target) will be positive integers.
- Elements in a combination (a1, a2,
… , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤
… ≤ ak). - The solution set must not contain duplicate combinations.
For example, given candidate set 2,3,6,7 and target 7,
A solution set is:
[7]
[2, 2, 3]
输出序列中和=target的种类，可重复使用
Combination Sum II
Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where
the candidate numbers sums to T.
Each number in C may only be used once in the combination.
Note:
- All numbers (including target) will be positive integers.
- Elements in a combination (a1, a2,
… , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤
… ≤ ak). - The solution set must not contain duplicate combinations.
For example, given candidate set 10,1,2,7,6,1,5 and target 8,
A solution set is:
[1, 7]
[1, 2, 5]
[2, 6]
[1, 1, 6]
同上题，加一个条件，出现的数只能使用一次
class Solution
{
vector<vector<int> >ans;
vector<int>num;
int key;
int len;
private:
void dfs(vector<int>mark,int n,int sum,int used)
{
if (sum==key)
{
ans.push_back(mark);
return ;
}
if (n==len) return ;
if (sum+num[n]>key) return ;
dfs(mark,n+1,sum,0);
if (num[n]==num[n-1] && used==0) return ;
mark.push_back(num[n]);
dfs(mark,n+1,sum+num[n],1);
mark.pop_back();
}
public:
vector<vector<int>> combinationSum2(vector<int>& candidates, int target)
{
len=candidates.size();
num=candidates;
sort(num.begin(),num.end());
key=target;
vector<int>mark;
dfs(mark,0,0,1);
return ans;
}
};
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