Description：

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 in ascending from left to right.
• Integers in each column are sorted in ascending from top to bottom.

 

For example,

Consider the following matrix:

[  [1,   4,  7, 11, 15],  [2,   5,  8, 12, 19],  [3,   6,  9, 16, 22],  [10, 13, 14, 17, 24],  [18, 21, 23, 26, 30]]

Given target = 5, return true.

Given target = 20, return false.

首先想到的就是遍历整个矩阵，时间复杂度是O(m*n)，肯定是Timeout。

public class Solution {    public boolean searchMatrix(int[][] matrix, int target) {                for(int i=0; i

然后在优化的话就想到了二分。对每一行进行二分。时间复杂度是O(n*logm)，还是Timeout，二分用的越多时间复杂度就越高，所以行列都二分（有点递归分治的意思）更会Timeout。

public class Solution {        public boolean binarySearch(int[] arr, int terget) {                int left = 0, int right = arr.length - 1;        while(left <= right) {            int mid = left + (right - left) / 2;            if(target == arr[mid]) {                return true;            }            if(target > arr[mid]) {                left = mid + 1;            }            else {                right = mid - 1;            }        }                return false;    }        public boolean searchMatrix(int[][] matrix, int target) {                for(int i=0; i

这么看来时间复杂度必须在线性的基础上才行。观察一下矩阵不难发现把矩阵逆时针旋转45度类似一棵二叉查找树。所以就可以模仿二叉查找树的方法来做了。

这样的话时间复杂度就是O(m + n)；AC。

public class Solution {            public boolean searchMatrix(int[][] matrix, int target) {                if(matrix.length==0 || matrix[0].length==0) {            return false;        }                int i = 0, j = matrix[0].length - 1;                while(i < matrix.length && j >= 0) {            int cur = matrix[i][j];            if(cur == target) {                return true;            }            else if(cur < target) {                i ++;            }            else {                j --;            }                    }        return false;            }    }