[LeetCode] Anagrams、N-Queens

最新推荐文章于 2020-06-07 23:13:40 发布

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本文深入探讨了Anagrams和N-Queens问题的解决策略，通过实例展示了如何使用C/C++语言实现这些复杂问题的高效解决方案。详细介绍了使用哈希表来快速匹配Anagrams的方法，以及解决N-Queens问题的经典回溯法，并提供了代码实现，旨在为读者提供实际应用中解决问题的实用技巧。

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Anagrams:

Given an array of strings, return all groups of strings that are anagrams.

Note: All inputs will be in lower-case.

比较Easy的题，用个map记录就是了。其实我可以说我没读懂题，百度人家的才懂吗。

class Solution {
public:
    vector<string> anagrams(vector<string> &strs) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        
        map<string,vector<string> > mask;
	    vector<string>::iterator it = strs.begin();
	    for(;it!=strs.end();++it)
	    {
		    string tmp=*it;
		    sort(tmp.begin(),tmp.end());
		    mask[tmp].push_back(*it);
	    }
	    vector<string> ans;
	    map<string,vector<string> >::iterator mit=mask.begin();
	    for(;mit!=mask.end();++mit)
	    {
		    if ( (*mit).second.size()>1)
		    {
			    it=(*mit).second.begin();
			    for(;it!=(*mit).second.end();++it)
				    ans.push_back(*it);
		    }
	    }
	    return ans;
    }
};

N-Queens：

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.."]
]

经典的Q皇后问题，注意左斜线和右斜线的下表即可。

#define vi vector<int>
#define vs vector<string>
#define vvs vector<vs >

class Solution {
public:
	vector<vector<string> > solveNQueens(int n) {
		// Start typing your C/C++ solution below
		// DO NOT write int main() function


		vvs ans;
		vs grid(n,string(n,'.'));
		vi col(n,0);
		vi L2R(2*n,0);
		vi R2L(2*n,0);
		solve(ans,grid,col,L2R,R2L,0);
		return ans;
	}
	void solve(vvs& ans,vs& grid, vi& col,vi& L2R,vi& R2L,int x)
	{
		int n = grid.size();
		if ( x>= n )
		{
			ans.push_back(grid);
			return;
		}

		for(int y=0;y<n;y++)
		{
			if ( col[y] || L2R[y-x+n-1] || R2L[x+y] )
				continue;
			col[y]=L2R[y-x+n-1]=R2L[x+y]=1;
			grid[x][y]='Q';
			solve(ans,grid,col,L2R,R2L,x+1);
			grid[x][y]='.';
			col[y]=L2R[y-x+n-1]=R2L[x+y]=0;
		}
	}
};