题目

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.

因为矩阵的行列都是递增的，所以可以利用左下角（或右上角）来与要找的数进行对比，然后对行列的标记进行增减即可，代码如下：

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

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