leetcode题解日练--2016.09.06

不给自己任何借口

今日题目：

1、数独的解

2、N-皇后II

3、N-皇后

今日摘录：

大抵浮生若梦，姑且此处销魂。
——曾国藩

37. Sudoku Solver | Difficulty: Hard

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.

tag：回溯|哈希表
题意：数独的解

思路：
1、dfs+回溯

class Solution {public:     void solveSudoku(vector<vector<char>>& board)     {        solve(board);     }    bool solve(vector<vector<char>>& board) {        for(int r = 0;r<9;r++)        {            for(int c=0;c<9;c++)            {                if(board[r][c]=='.')                {                    for (char k = '1'; k <= '9'; k++)                    {                        if(isvalid(board,r,c,k))                        {                            board[r][c] = k;                            if(solve(board))    return true;                            board[r][c] = '.';                        }                    }                    return false;                }            }        }        return true;    }    bool isvalid(vector<vector<char>>& board,int row,int col,char k)    {        for(int i=0;i<9  ;i++)            if(board[i][col]==k)    return false;        for(int i=0;i<9 ;i++)            if(board[row][i]==k)    return false;        for(int i=(row/3)*3;i<(row/3+1)*3 ;i++)            {                for(int j=(col/3)*3;j<(col/3+1)*3;j++)                {                    if(board[i][j]==k)    return false;                }            }        return true;    }};

结果：53ms

52. N-Queens II | Difficulty: Hard

Follow up for N-Queens problem.

Now, instead outputting board configurations, return the total number of distinct solutions.

tag：回溯

题意：找到N皇后有多少个解
思路：
1、经典的回溯思想，边放置边剪枝。

class Solution {public:int totalNQueens(int n) {    int res = 0;    //cols代表因为列不冲突的情况，main是主对角线，anti是辅对角线。为何对角线的标志位数组要比cols大呢？    vector<bool> cols(n,true);    vector<bool> main(2*n-1,true);    vector<bool> anti(2*n-1,true);    dfs(0,res,cols,main,anti);    return res;}void dfs(int row,int&res,vector<bool> &cols,vector<bool> &main,vector<bool> &anti){    if(row==cols.size())    {        res++;        return ;    }     for(int col = 0;col<cols.size();col++)    {        if(cols[col] && main[row-col+cols.size()-1]&&anti[row+col])        {            cols[col] = main[row-col+cols.size()-1]=anti[row+col] = false;            dfs(row+1,res,cols,main,anti);            cols[col] = main[row-col+cols.size()-1]=anti[row+col] = true;        }    }    return;}};

结果：13ms

2、
使用位运算的方法：
row用一个Nbit位的数代表哪些位置是接下来还能放的，哪些位置是不能放的。
main：主对角线，代表主对角线对下一行选取位置的影响
anti：副对角线，代表副对角线对下一行选取位置的影响

class Solution {public:void queenBit(int row,int main,int anti,const int all_queen){    if(row!=all_queen)    {        int pos = all_queen & ~(row|main|anti);        while(pos!=0)        {            int p = pos&-pos;            pos-=p;            queenBit(row+p,(main+p)<<1,(anti+p)>>1,all_queen);        }    }    else    {        res++;        return ;    }}int totalNQueens(int n) {    res = 0;    int all_queen = (1<<n)-1;    queenBit(0,0,0,all_queen);    return res;}private:    int res;};

结果：0ms

51. N-Queens | Difficulty: Hard

The n-queens puzzle is the problem of placing n queens on an n×n chessboard such that no two queens attack each other.

Given an integer n, return all distinct solutions to the n-queens puzzle.

Each solution contains a distinct board configuration of the n-queens’ placement, where ‘Q’ and ‘.’ both indicate a queen and an empty space respectively.

For example,
There exist two distinct solutions to the 4-queens puzzle:

[
[“.Q..”, // Solution 1
“…Q”,
“Q…”,
“..Q.”],

[“..Q.”, // Solution 2
“Q…”,
“…Q”,
“.Q..”]
]
tag：回溯

题意：找到N皇后的每个解是什么。
思路：
1、经典的回溯思想，边放置边剪枝。

class Solution {public:    void dfs(int row,vector<vector<string>>&res,vector<string>path,vector<bool> &cols,vector<bool> &main,vector<bool> &anti,const string str)    {        if(row==cols.size())        {            res.push_back(path);            return ;        }        for(int col = 0;col<cols.size();col++)        {            if(cols[col] && main[row+col]&&anti[row-col+cols.size()-1])            {                string tmp = str;                tmp[col] = 'Q';                path.push_back(tmp);                cols[col] = main[row+col]=anti[row-col+cols.size()-1] = false;                dfs(row+1,res,path,cols,main,anti,str);                path.pop_back();                tmp[col] = '.';                cols[col] = main[row+col]=anti[row-col+cols.size()-1] = true;            }        }        return;    }    vector<vector<string>> solveNQueens(int n) {        vector<bool> cols(n,true);        vector<bool> main(2*n-1,true);        vector<bool> anti(2*n-1,true);        vector<string> path;        string str = dot.substr(0,n);         dfs(0,res,path,cols,main,anti,str);        return res;    }private:    vector<vector<string>>res;    string dot = "......................";};

结果：29ms

2、
使用位运算的方法：
row用一个Nbit位的数代表哪些位置是接下来还能放的，哪些位置是不能放的。
main：主对角线，代表主对角线对下一行选取位置的影响
anti：副对角线，代表副对角线对下一行选取位置的影响

class Solution {public:    void queenBit(int row,int main,int anti,const int all_queen,const string str,vector<string>&path)    {        if(row!=all_queen)        {            int pos = all_queen & ~(row|main|anti);            while(pos!=0)            {                int p = pos&-pos;                pos-=p;                string tmp = str;                int i = log10(p)/log10(2);                tmp[i] = 'Q';                path.push_back(tmp);                queenBit(row+p,(main+p)<<1,(anti+p)>>1,all_queen,str,path);                path.pop_back();                tmp[i] = '.';            }        }        else        {            res.push_back(path);            return ;        }    }    vector<vector<string>> solveNQueens(int n) {        int all_queen = (1<<n)-1;        string str = dot.substr(0,n);        vector<string> path;        queenBit(0,0,0,all_queen,str,path);        return res;    }private:    vector<vector<string>>res;    string dot = "......................";};

结果：3ms