(medium)LeetCode 240.Search a 2D Matrix II

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.

解法一：暴力解法，遍历二维数组。

代码如下：

   

public class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        int rows=matrix.length;
        int columns=matrix[0].length;
        for(int i=0;i<rows;i++)
          for(int j=0;j<columns;j++){
              if(target==matrix[i][j])
                 return true;
          }
        return false;  
    }
}

　　运行结果：时间复杂度为O(M*N)

   

方法二：根据矩阵的特点，从右上角元素开始比较，target小，则列数减1，target大，则行数加1.

代码如下：

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

　　运行结果：时间复杂度O(M+N)