leetcode 题解代码整理 36-40题

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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.

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;
    }
};

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.

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 1" or 11.
11 is read off as "two 1s" or 21.
21 is read off as "one 2, then one 1" or 1211.

Given an integer n, generate the nth sequence.

Note: The sequence of integers will be represented as a string.

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;

    }
};

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] 

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] 

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;
}
};

leetcode 题解代码整理 36-40题

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原文地址：http://blog.csdn.net/u011932355/article/details/47790213