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。

 1 class Solution {
 2     class Node {
 3         int i, j;
 4         int v;
 5         public Node(int i, int j, int v) {
 6             this.i = i;
 7             this.j = j;
 8             this.v = v;
 9         }
10     }
11     
12     static Comparator<Node> cmp = new Comparator<Node>() {
13         public int compare(Node e1, Node e2) {
14             return e1.v - e2.v;
15         }
16     };
17     
18     boolean find(int []value, int target) {
19         int n = value.length;
20         int l = 0, r = n - 1, mid;
21         while (l <= r) {
22             mid = (l + r) / 2;
23             if (value[mid] > target) {
24                 r = mid - 1;
25             } else if (value[mid] < target) {
26                 l = mid + 1;
27             } else {
28                 return true;
29             }
30         }
31         
32         return false;
33         
34     }
35    
36     public boolean searchMatrix(int[][] matrix, int target) {
37         if (matrix.length == 0 || matrix[0].length == 0) return false;
38         Queue<Node> queue = new PriorityQueue<Node>(cmp);
39         queue.add(new Node(0, 0, matrix[0][0]));
40         int [] value = new int[matrix.length * matrix[0].length];
41         boolean [][] flag = new boolean[matrix.length][matrix[0].length];
42         int k = 0;
43         while (!queue.isEmpty()) {
44             Node temp = queue.poll();
45             value[k++] = temp.v;
46             if (temp.i + 1 < matrix.length && !flag[temp.i + 1][temp.j]) {
47                 queue.add(new Node(temp.i + 1, temp.j, matrix[temp.i + 1][temp.j]));
48                 flag[temp.i + 1][temp.j] = true;
49             }
50             if (temp.j + 1 < matrix[0].length && !flag[temp.i][temp.j + 1] ) {
51                 queue.add(new Node(temp.i, temp.j + 1, matrix[temp.i][temp.j + 1]));
52                 flag[temp.i][temp.j + 1] = true;
53             }
54         }
55         return find(value, target);
56     }
57 }

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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

 1 class Solution {
 2    
 3     public boolean searchMatrix(int[][] matrix, int target) {
 4         if (matrix.length == 0 || matrix[0].length == 0) return false;
 5         int m = matrix.length, n = matrix[0].length;
 6         int i = 0, j = n - 1;
 7         while (i < m && j >= 0) {
 8             if (matrix[i][j] > target) {
 9                 j--;
10             } else if (matrix[i][j] < target) {
11                 i++;
12             } else {
13                 return true;
14             }
15         }
16         return false;
17     }
18 }

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