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The problem:
Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:
For example,
Consider the following matrix:
[ [1, 3, 5, 7], [10, 11, 16, 20], [23, 30, 34, 50] ]
Given target = 3
, return true
.
Solution1:
public class Solution { public boolean searchMatrix(int[][] matrix, int target) { if (matrix == null || matrix.length == 0) //sterotype checking return false; int low = 0; int high = matrix.length - 1; int mid = -1; int row_index = -1; while (low <= high) { //search the row that may contain target mid = (low + high) / 2; if (matrix[mid][0] == target) { return true; } else if (matrix[mid][0] < target) { low = mid + 1; } else { high = mid - 1; } } /*a little skill: when exit while loop 1. array[low] is the element just bigger than target; 2. array[high] is th element just smaller than target. if (matrix[mid][0] > target) // the row that may contain target (mid + 1 is the possible to insert) row_index = mid - 1; else row_index = mid; */ row_index = high; if (row_index < 0) //the target is even smaller than the smallest element in the matrix(first element) return false; low = 0; high = matrix[0].length - 1; mid = -1; while (low <= high) { //serach the target in the possible row mid = (low + high) / 2; if (matrix[row_index][mid] == target) { return true; } else if (matrix[row_index][mid] < target) { low = mid + 1; } else { high = mid - 1; } } return false; } }
Solution2:
public class Solution { public boolean searchMatrix(int[][] matrix, int target) { int row_count = matrix.length; int column_count = matrix[0].length; if (matrix == null || matrix[0][0] > target || matrix[row_count-1][column_count-1] < target) return false; int i = 0; int j = column_count - 1; while (i <= row_count-1 && j >= 0) { if (matrix[i][j] == target) return true; else if (matrix[i][j] < target) i++; else j--; } return false; } }
[LeetCode#74]Search a 2D Matrix
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原文地址:http://www.cnblogs.com/airwindow/p/4302505.html