[LintCode] Search a 2D Matrix
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Write an efficient algorithm that searches for a value in an m x 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.
4/23/2017
算法班
corner case想的不够全面,第23-28行是后来加上去的,原因是如果target在最后一行或者比任何一个值都大,我们是需要在end那一行找答案的。其他所有情况都是在start那一行找。
1 public class Solution { 2 /** 3 * @param matrix, a list of lists of integers 4 * @param target, an integer 5 * @return a boolean, indicate whether matrix contains target 6 */ 7 public boolean searchMatrix(int[][] matrix, int target) { 8 // write your code here 9 10 if (matrix == null || matrix.length == 0 || matrix[0].length == 0) return false; 11 boolean ret; 12 int start = 0, end = matrix.length - 1; 13 14 while (start + 1 < end) { 15 int mid = start + (end - start) / 2; 16 if (matrix[mid][0] == target) return true; 17 if (matrix[mid][0] < target) { 18 start = mid; 19 } else { 20 end = mid; 21 } 22 } 23 int row; 24 if (matrix[end][0] < target) { 25 row = end; 26 } else { 27 row = start; 28 } 29 int left = 0, right = matrix[row].length - 1; 30 while (left + 1 < right) { 31 int mid = left + (right - left) / 2; 32 if (matrix[row][mid] == target) return true; 33 if (matrix[row][mid] < target) { 34 left = mid; 35 } else { 36 right = mid; 37 } 38 } 39 if (matrix[row][right] == target || matrix[row][left] == target) return true; 40 return false; 41 } 42 }
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