Search a 2D Matrix ——LeetCode

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.

 

For example,

Consider the following matrix:

[

  [1,   3,  5,  7],

  [10, 11, 16, 20],

  [23, 30, 34, 50]

]

Given target = 3, return true.

题目大意：给一个m*n矩阵，每一行都是递增有序，逐行也是递增的，要求设计一个高效算法检查目标元素是否存在于此矩阵中。

解题思路：可以把整个矩阵展开，就是一个长数组，数组长度为M*N，用二分查找即可，就是需要把二分查找的位置转化为矩阵下标。假设有row行，col列，那么key对应的矩阵中的元素应该是matrix[key/col][key%col]，这样就转为二分查找了。

Talk is cheap>>

 　　public boolean searchMatrix(int[][] matrix, int target) {

        if (matrix == null || matrix[0][0] > target)

            return false;

        int rowLen = matrix.length;

        int colLen = matrix[0].length;

        int low = 0, high = rowLen * colLen - 1;



        while (low <= high) {

            int mid = (low + high) >>> 1;

            int x = mid / colLen;

            int y = mid % colLen;

            if (matrix[x][y] > target) {

                high = mid - 1;

            } else if (matrix[x][y] < target) {

                low = mid + 1;

            } else {

                return true;

            }

        }

        return false;

    }