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.

思路:
这道题非常有技巧,我们选择右上角的那个点,然后进行比较,如果大,则向下走,如果小则向右。(因为从那个点开始,下面永远大于,左边永远小于)

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

喜欢的话订阅一个呗~第一时间收到文章更新哟~

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