Follow up for N-Queens problem.

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

和N-Queens 同样的解法。<??http://www.2cto.com/kf/ware/vc/" target="_blank" class="keylink">vcD4NCjxwcmUgY2xhc3M9"brush:java;"> class Solution { public: int totalNQueens(int n) { vector > result; vector nums; for(int i=0;i > base=permutation(nums); return base.size(); } vector > permutation(vector &nums) { vector >result; permutationChild(nums,0,result); return result; } void permutationChild(vector &nums,int offset,vector >&result) { if(offset==nums.size()-1) { if(judge(nums)) result.push_back(nums); return; } auto base=nums.begin()+offset; auto iter=base; for(;iter!=nums.end();iter++) { swap(*base,*iter); permutationChild(nums,offset+1,result); swap(*base,*iter); } } bool judge(vector & nums) { int length=nums.size(); for(int i=0;i

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