LeetCode 74. Search a 2D Matrix

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.

在一个有序的二维数组里面查找一个值，思路是先对行二分查找，再对这行二分查找，不过我的代码只二分查找了行，遍历了这一行也能AC。。

public class Solution {    public boolean searchMatrix(int[][] matrix, int target) {        if(matrix.length==0||matrix[0].length==0)return false;        int row = matrix.length-1;        int col = matrix[0].length-1;        int l = 0;        int r = row;        int m = (l+r)/2;        int c = 0;        while(l<=r){            if(matrix[m][0]>target&&m!=0&&matrix[m-1][0]<=target){                c = m-1;                break;            }            if(matrix[m][col]<target&&m!=row&&matrix[m+1][col]>=target){                c = m+1;                break;            }            if(matrix[m][0]>target){                r = m-1;                m = (l+r)/2;            }            else{                l = m+1;                m = (l+r)/2;            }        }        for(int i=0;i<=col;i++){            if(matrix[c][i]==target)return true;        }        return false;    }}