28. Search a 2D Matrix 【easy】

28. Search a 2D Matrix 【easy】

Write an efficient algorithm that searches for a value in an mx 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.

Example

Consider the following matrix:

[
    [1, 3, 5, 7],
    [10, 11, 16, 20],
    [23, 30, 34, 50]
]

Given target = 3, return true.

Challenge

O(log(n) + log(m)) time

解法一：

 class Solution {
 public:
     /*
      * @param matrix: matrix, a list of lists of integers
      * @param target: An integer
      * @return: a boolean, indicate whether matrix contains target
      */
     bool searchMatrix(vector<vector<int>> &matrix, int target) {
         int row = matrix.size();
         if (row == ) {
             return false;
         }

         int col = matrix[].size();
         if (col == ) {
             return false;
         }

         int start = ;
         int end = row * col - ;

         while (start +  < end) {
             int mid = start + (end - start) / ;

             if (matrix[mid / col][mid % col] == target) {
                 return true;
             }
             else if (matrix[mid / col][mid % col] < target) {
                 start = mid;
             }
             else if (matrix[mid / col][mid % col] > target) {
                 end = mid;
             }
         }

         if (matrix[start / col][start % col] == target
             || matrix[end / col][end % col] == target) {
             return true;
         }
         else {
             return false;
         }
     }
 };

注意：

1、边界值合法性检查

2、二维数组转一维数组的计算方式

解法二：

 public class Solution {
     public boolean searchMatrix(int[][] matrix, int target) {
         if (matrix == null || matrix.length == ) {
             return false;
         }
         if (matrix[] == null || matrix[].length == ) {
             return false;
         }

         int row = matrix.length;
         int column = matrix[].length;

         // find the row index, the last number <= target
         int start = , end = row - ;
         while (start +  < end) {
             int mid = start + (end - start) / ;
             if (matrix[mid][] == target) {
                 return true;
             } else if (matrix[mid][] < target) {
                 start = mid;
             } else {
                 end = mid;
             }
         }
         if (matrix[end][] <= target) {
             row = end;
         } else if (matrix[start][] <= target) {
             row = start;
         } else {
             return false;
         }

         // find the column index, the number equal to target
         start = ;
         end = column - ;
         while (start +  < end) {
             int mid = start + (end - start) / ;
             if (matrix[row][mid] == target) {
                 return true;
             } else if (matrix[row][mid] < target) {
                 start = mid;
             } else {
                 end = mid;
             }
         }
         if (matrix[row][start] == target) {
             return true;
         } else if (matrix[row][end] == target) {
             return true;
         }
         return false;
     }
 }

两次二分，值得借鉴！