Leetcode 240. Search a 2D Matrix II

题目链接: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.

  此题是74题Search a 2D Matrix的升级版,所给出的矩阵性质相对74题少了一条,只保证了每行和每列都是增序的,但依旧有O(m+n)的解法。

  具体思路就是每一行倒着扫,扫到第一个比target小的数就跳到下行,如果等于当然是直接返回true了,如果下一行还比target小就继续跳下一行,直到最后一行。

  为啥这么做是可行的? 可能我比较笨,想了半天才想到。 因为每一列都是增序的,举个例子,假设matrix[0][5] > target,那么[0][5]位置右下(包含右和下)所有元素不可能比target小。

直接上代码

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