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. Challenge O(m+n) time and O(1) extra space
很巧妙的思路，可以从左下或者右上开始找
1 public class Solution { 2 /** 3 * @param matrix: A list of lists of integers 4 * @param: A number you want to search in the matrix 5 * @return: An integer indicate the occurrence of target in the given matrix 6 */ 7 public int searchMatrix(int[][] matrix, int target) { 8 // write your code here 9 if (matrix==null || matrix.length==0 || matrix[0].length==0) return 0; 10 int m = matrix.length; 11 int n = matrix[0].length; 12 int count = 0; 13 int row = m-1; 14 int col = 0; 15 while (row>=0 && row<m && col>=0 && col<n) { 16 int cur = matrix[row][col]; 17 if (cur == target) { 18 count++; 19 col++; 20 row--; 21 } 22 else if (cur > target) { 23 row--; 24 } 25 else col++; 26 } 27 return count; 28 } 29 }
这道题的一个优化是对于一个矩阵的最后一行做二分搜索后，删掉前几列和最后一行，得到一个子矩阵。重复这样的操作，时间复杂度是O(min(m,n)log(max(m,n)))。之后跟她提了一下这个方法在m和n相差比较大的时候可能比较有用。
2, 20, 35, 41, 70,
11, 24, 44, 60, 78,
21, 30, 50, 75, 90,
31, 38, 99, 100, 102