51. N 皇后

class Solution {
    List<List<String>> res = new ArrayList<>();
    public List<List<String>> solveNQueens(int n) {
        if (n<=0) return res;
        char[][] board = new char[n][n];
        for (char[] chars : board) Arrays.fill(chars,'.');
        backtrack(board,0);
        return res;
    }

    public void backtrack(char[][] board,int row){
        if(board.length==row){
            res.add(charToStr(board));
            return;
        }
        int n = board[row].length;
        for(int col = 0;col < n;col++){
            if(!isValid(board,row,col)) continue;
            board[row][col] = 'Q';
            backtrack(board,row+1);
            board[row][col] = '.';
        }
    }

    public List<String> charToStr(char[][] board){
        List<String> result = new ArrayList<>();
        for (char[] c : board){
            result.add(String.valueOf(c));
        }
        return result;
    }

    public boolean isValid(char[][] board,int row,int col){
        int rows= board.length;
        for (char[] c : board){
            if (c[col]=='Q'){
                return false;
            }
        }
        // 左上角
        for (int i = row-1,j = col-1;i >=0 && j >= 0;i--,j--){
            if (board[i][j]=='Q') return false;
        }
        // 右上角
        for (int i = row-1,j = col+1;i >=0 && j < rows;i--,j++){
            if (board[i][j]=='Q') return false;
        }
        return true;
    }
}

劍指 Offer 20. 表示數值的字串

class Solution {
    public boolean isNumber(String s) {
        if (s.length()==0 || s == null){
            return false;
        }
        s = s.trim();
        boolean numFlag = false;
        boolean dotFlag = false;
        boolean eFlag = false;
        for (int i = 0; i < s.length(); i++) {
            char c = s.charAt(i);
            // 判斷為數字  numFlag 標記為true
            if (c >= '0' && c <= '9'){
                numFlag = true;
                // 判斷為 . 時，之前需要沒有出現過 . 和 e
            }else if (c=='.' && !dotFlag && !eFlag){
                dotFlag = true;
                // 判斷為e時，之前需要出現過數字，但是沒有出現過e
            }else if ((c=='e' || c=='E') && numFlag && !eFlag){
                eFlag = true;
                numFlag = false;
            }else if ((c=='+' || c=='-') && (i==0 || s.charAt(i-1)=='e' || s.charAt(i-1)=='E')){

            }else {
                return false;
            }
        }
        return numFlag;
    }
}

劍指 Offer 41. 資料流中的中位數

class MedianFinder {
    PriorityQueue<Integer> left;
    PriorityQueue<Integer> right;
    /** initialize your data structure here. */
    public MedianFinder() {
        left = new PriorityQueue<>((n1,n2)->n2-n1);
        right = new PriorityQueue<>();
    }
    
    public void addNum(int num) {
        left.add(num);
        right.add(left.poll());
        if(left.size()+1 < right.size()){
            left.add(right.poll());
        }
    }
    
    public double findMedian() {
        if(right.size()>left.size())return right.peek();
        return (left.peek()+right.peek())/2.0;
    }
}

class MedianFinder {
    Queue<Integer> A, B;
    public MedianFinder() {
        A = new PriorityQueue<>(); // 小頂堆，儲存較大的一半
        B = new PriorityQueue<>((x, y) -> (y - x)); // 大頂堆，儲存較小的一半
    }
    public void addNum(int num) {
        if(A.size() != B.size()) {
            A.add(num);
            B.add(A.poll());
        } else {
            B.add(num);
            A.add(B.poll());
        }
    }
    public double findMedian() {
        return A.size() != B.size() ? A.peek() : (A.peek() + B.peek()) / 2.0;
    }
}

你知道的越多，你不知道的越多。