leetcode hard模式专杀之51. N-Queens

继续leetcode hard模式， 八皇后问题的一般版，这种举一反三的题目值得好好琢磨，典型的回溯法，思路就不多说了，上代码，注意一一些边界条件的细节：

public class Solution {    public List<String>  transfer(int[] q){        List<String> result = new ArrayList<>();        for(int i = 0; i<q.length; i++){            StringBuilder sb = new StringBuilder();            for(int j=0; j<q.length; j++){                if(j==q[i]){                    sb.append("Q");                }else{                    sb.append(".");                }            }            result.add(sb.toString());        }        return result;    }    public List<List<String>> solveNQueens(int n) {        List<List<String>> result = new ArrayList<>();        int[] q=new int[n];        for(int i = 0 ; i<n; i++){            enumerate(result, i, q, 0);        }        return result;    }    public boolean isConsistent(int[] q, int line, int index){        if(line==0){            return true;        }else{            for(int i=0; i<line;i++){                //i行q[i]列 vs line行index列                if(q[i]==index){                    //同列                    return false;                }                if(index-q[i]==line-i){                    return false;                }                if(index-q[i]==i-line){                    return false;                }            }            return true;        }    }    public void enumerate(List<List<String>> result, int index, int[] q, int line){        if(line==q.length-1){            if(isConsistent(q, line, index)){                q[line] = index;                result.add(transfer(q));            }        }else if(line == 0){            q[line] = index;            for(int j = 0; j<q.length; j++){                enumerate(result, j, q, line+1);            }        }else{            if(isConsistent(q, line, index)){                q[line] = index;                for(int j = 0; j<q.length; j++){                    enumerate(result, j, q, line+1);                }            }        }    }}