Dynamic Programming - Eight Queens

We do a lot of dynamic programming topics, with some experience.

    public int getanswer(char[][] map,int index,int n) {
        if(index==n) {// index==n  则意味的递归结束
            /*System.out.println("-------------");
            for(int i=0;i<n;i++) {
                System.out.println(map[i]);
            }//打印一些 可行的排列
            System.out.println("--------------");*/
            return -1;//返回-1 意味着递归的结束，或 递归失败
        }else {
            for(int i=0;i<n;i++) {//遍历该次 皇后可行的位置
                if(isOk(i, index, map)) {
                    map[index][i]='Q';
                    if(getanswer(map, index+1, n)==-1) {//递归返回1  说明该位置符合
                        map[index][i]='\0';
                    }
                }
            }
            return -1;//递归失败，返回上一级
        }
    }
    public boolean isOk(int x,int y,char[][] map) {//判断某个点 是否可以安放
        int len=map.length;
        for(int i=y-1,j=1;i>=0;i--) {
            if(map[i][x]!='\0')return false;
            if(x-j>=0&&map[i][x-j]!='\0')return false;
            if(x+j<len&&map[i][x+j]!='\0')return false;
            j++;
        }
        return true;
    }

Overall, the eight queens problem is in constant recursive solution, the attempt possible, if the next level of recursion fails, try case