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:
Consider the following matrix:
[
[1, 3, 5, 7],
[2, 4, 7, 8],
[3, 5, 9, 10]
]
Given target = 3, return 2.
O(m+n) time and O(1) extra space
解题思路：
从左下角往右上角找，若是小于target就往右找，若是大于target就往上找。时间复杂度O(m+n) n 为行数，m为列数。
定义count 计数。
与 Leetcode 240. Search a 2D Matrix II 类似。
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) { //check corner case if (matrix == null || matrix.length == 0) { return 0; } if (matrix[0] == null || matrix[0].length == 0) { return 0; } //find from bottom left to top right int n = matrix.length; //row int m = matrix[0].length; //column int x = n - 1; int y = 0; int count = 0; while (x >= 0 && y < m) { if (matrix[x][y] < target) { y++; } else if (matrix[x][y] > target) { x--; } else { count++; x--; y++; } } return count; } }
Reference:
1. http://www.jiuzhang.com/solutions/search-a-2d-matrix-ii/
LintCode 38. Search a 2D Matrix II
原文：http://www.cnblogs.com/anne-vista/p/5135401.html