【Leetcode】51. N-Queens(回溯)
2016-09-04 11:11
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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
indicate a queen and an empty space respectively.
For example,
There exist two distinct solutions to the 4-queens puzzle:
class Solution {
private:
vector<vector<string>> key;
public:
vector<vector<string>> solveNQueens(int n) {
vector<int> vi(n);
solveNQueens(vi,0,n);
return key;
}
bool isSuit(int curIndex, int val ,int n, vector<int> &a){
if(curIndex<n){
for(int i=0;i<curIndex;i++){
if(a[i]==val||(a[i]-val)==(i-curIndex)||(a[i]-val)==(curIndex-i)) return false;
}
return true;
}
else
return false;
}
void solveNQueens(vector<int> &vi,int c,int n) {
if(c==n){
vector<string> s(n,string(n,'.'));
for(int i=0;i<n;i++){
s[i][vi[i]]='Q';
}
key.push_back(s);
return;
}
for(vi[c] = 0; vi[c] < n; ++ vi[c])
if(isSuit(c,vi[c],n,vi))
solveNQueens(vi, c + 1,n);
}
};
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.."] ]
class Solution {
private:
vector<vector<string>> key;
public:
vector<vector<string>> solveNQueens(int n) {
vector<int> vi(n);
solveNQueens(vi,0,n);
return key;
}
bool isSuit(int curIndex, int val ,int n, vector<int> &a){
if(curIndex<n){
for(int i=0;i<curIndex;i++){
if(a[i]==val||(a[i]-val)==(i-curIndex)||(a[i]-val)==(curIndex-i)) return false;
}
return true;
}
else
return false;
}
void solveNQueens(vector<int> &vi,int c,int n) {
if(c==n){
vector<string> s(n,string(n,'.'));
for(int i=0;i<n;i++){
s[i][vi[i]]='Q';
}
key.push_back(s);
return;
}
for(vi[c] = 0; vi[c] < n; ++ vi[c])
if(isSuit(c,vi[c],n,vi))
solveNQueens(vi, c + 1,n);
}
};
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