240. Search a 2D Matrix II

Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:

• Integers in each row are sorted in ascending from left to right.
• Integers in each column are sorted in ascending from top to bottom.

Example:

Consider the following matrix:

[
  [1,   4,  7, 11, 15],
  [2,   5,  8, 12, 19],
  [3,   6,  9, 16, 22],
  [10, 13, 14, 17, 24],
  [18, 21, 23, 26, 30]
]

Given target = 5, return true.

Given target = 20, return false.

第一种方法：一下子没有想到好的方法，只利用了行已经排过序这个性质，每行进行一次二分查找。

class Solution {
    class Node {
        int i, j;
        int v;
        public Node(int i, int j, int v) {
            this.i = i;
            this.j = j;
            this.v = v;
        }
    }
    
    static Comparator<Node> cmp = new Comparator<Node>() {
        public int compare(Node e1, Node e2) {
            return e1.v - e2.v;
        }
    };
    
    boolean find(int []value, int target) {
        int n = value.length;
        int l = 0, r = n - 1, mid;
        while (l <= r) {
            mid = (l + r) / 2;
            if (value[mid] > target) {
                r = mid - 1;
            } else if (value[mid] < target) {
                l = mid + 1;
            } else {
                return true;
            }
        }
        
        return false;
        
    }
   
    public boolean searchMatrix(int[][] matrix, int target) {
        if (matrix.length == 0 || matrix[0].length == 0) return false;
        Queue<Node> queue = new PriorityQueue<Node>(cmp);
        queue.add(new Node(0, 0, matrix[0][0]));
        int [] value = new int[matrix.length * matrix[0].length];
        boolean [][] flag = new boolean[matrix.length][matrix[0].length];
        int k = 0;
        while (!queue.isEmpty()) {
            Node temp = queue.poll();
            value[k++] = temp.v;
            if (temp.i + 1 < matrix.length && !flag[temp.i + 1][temp.j]) {
                queue.add(new Node(temp.i + 1, temp.j, matrix[temp.i + 1][temp.j]));
                flag[temp.i + 1][temp.j] = true;
            }
            if (temp.j + 1 < matrix[0].length && !flag[temp.i][temp.j + 1] ) {
                queue.add(new Node(temp.i, temp.j + 1, matrix[temp.i][temp.j + 1]));
                flag[temp.i][temp.j + 1] = true;
            }
        }
        return find(value, target);
    }
}

 第二种方法：每次都从矩阵右上角的元素v开始，如果target > v则该元素所在行肯定不存在target，从矩阵删除该行， 如果target < v 则该元素所在列肯定不存在target, 从矩阵删除该列，如果target = v 则找到该元素

class Solution {
   
    public boolean searchMatrix(int[][] matrix, int target) {
        if (matrix.length == 0 || matrix[0].length == 0) return false;
        int m = matrix.length, n = matrix[0].length;
        int i = 0, j = n - 1;
        while (i < m && j >= 0) {
            if (matrix[i][j] > target) {
                j--;
            } else if (matrix[i][j] < target) {
                i++;
            } else {
                return true;
            }
        }
        return false;
    }
}