LeetCode 240 Search a 2D Matrix II

题目描述

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.

分析

参考《剑指offer》,每一次排除一行或一列。

《LeetCode 240 Search a 2D Matrix II》

查找数字7:

  1. 首先寻找matrix的左上角的元素9, 9>7,所以排除掉9所在的列,因为9以下的列,肯定比9大
  2. 再次寻找剩余matrix的左上角的元素8, 8>7,所以排除掉8所在的列
  3. 再次寻找剩余matrix的左上角的元素2, 2<7,所以排除掉2所在的行,因为2所在行,之前的元素都比2小
  4. 依次递推

代码

    public boolean searchMatrix(int[][] matrix, int target) {

        if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
            return false;
        }

        int m = matrix.length;
        int n = matrix[0].length;

        int x = 0, y = n - 1;

        while (x < m && y >= 0) {
            if (matrix[x][y] == target) {
                return true;
            } else if (matrix[x][y] < target) {
                x++;
            } else {
                y--;
            }
        }

        return false;
    }
    原文作者:_我们的存在
    原文地址: https://blog.csdn.net/Yano_nankai/article/details/50273129
    本文转自网络文章,转载此文章仅为分享知识,如有侵权,请联系博主进行删除。
点赞