38. Search a 2D Matrix II [LintCode]

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, return2.

对比第一题，本题每一行有overlap，所以不能拉成一个序列进行二分搜索。但是可以针对每一行或者每一列单独进行二分搜索，复杂度为O(m*logn)或O(n*logm)。

还可以用走对角线的方法，因为在垂直和水平方向分别是递增的，所以当一个元素大于或者小于target的时候，可以排除掉一行或者一列（x++, y--)，复杂度为O(m + n)

public class Solution {
    /**
     * @param matrix: A list of lists of integers
     * @param: A number you want to search in the matrix
     * @return: An integer indicate the occurrence of target in the given matrix
     */
    public int searchMatrix(int[][] matrix, int target) {
        // write your code here
        if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
            return 0;
        }
        int count = 0;
        int x = 0;
        int y = matrix[0].length - 1;
        while (x < matrix.length && y >= 0) {
            if (matrix[x][y] < target) {
                x++;
            } else if (matrix[x][y] > target) {
                y--;
            } else {
                count++;
                y--;
                x++;
            }
        }
        return count;
    }
}