Search a 2D Matrix | & II

Search a 2D Matrix II

Write an efficient algorithm that searches for a value in an m x n matrix, return the occurrence of it.

This matrix has the following properties:

• Integers in each row are sorted from left to right.
• Integers in each column are sorted from up to bottom.
• No duplicate integers in each row or column.

Example

Consider the following matrix:

[
  [1, 3, 5, 7],
  [2, 4, 7, 8],
  [3, 5, 9, 10]
]

Given target = 3, return 2.

分析：

因为数组里的数有上面三个特性，所以我们可以从左下角开始找。如果当前值比target大，明显往上走，比target小，往右走，如果一样斜上走。Time complexity: (O(m + n)) (m = matrix.length, n = matrix[0].length)

 public class Solution {
     public int searchMatrix(int[][] matrix, int target) {
         // check corner case
         if (matrix == null || matrix.length == 0 || matrix[].length == 0) {
             return ;
         }
         // from bottom left to top right
         int x = matrix.length - ;
         int y = ;
         int count = ;

         while (x >=  && y < matrix[].length) {
             if (matrix[x][y] < target) {
                 y++;
             } else if (matrix[x][y] > target) {
                 x--;
             } else {
                 count++;
                 x--;
                 y++;
             }
         }
         return count;

     }
 }

Search a 2D Matrix I

Write an efficient algorithm that searches for a value in an mx n matrix.

This matrix has the following properties:

• Integers in each row are sorted from left to right.
• The first integer of each row is greater than the last integer of the previous row.

Example

Consider the following matrix:

[
    [1, 3, 5, 7],
    [10, 11, 16, 20],
    [23, 30, 34, 50]
]

Given target = 3, return true.

分析：

根据数组的特点，我们需要找出一个大范围（row），然后再确定小范围。

 public class Solution {
     public boolean searchMatrix(int[][] matrix, int target) {
         if (matrix == null || matrix.length ==  || matrix[].length == ) return false;
         int rowCount = matrix.length, colCount = matrix[].length;
         if (matrix[][] > target || matrix[rowCount - ][colCount - ] < target) return false;

         int row = getRowNumber(matrix, target);
         int column = getColumnNumber(matrix, row, target);

         return matrix[row][column] == target;

     }

     public int getRowNumber(int[][] matrix, int target) {
         int start = , end = matrix.length - , width = matrix[].length;
         while (start <= end) {
             int mid = start + (end - start) / ;
             if (matrix[mid][width - ] == target) {
                 return mid;
             } else if (matrix[mid][width - ] > target) {
                 end = mid - ;
             } else {
                 start = mid + ;
             }
         }
         return start;
     }

     public int getColumnNumber(int[][] matrix, int row, int target) {
         int start = , end = matrix[].length;

         while (start <= end) {
             int mid = start + (end - start) / ;
             if (matrix[row][mid] == target) {
                 return mid;
             } else if (matrix[row][mid] > target) {
                 end = mid - ;
             } else {
                 start = mid + ;
             }
         }
         return start;
     }
 }