zoukankan      html  css  js  c++  java
  • 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.

    Analyse: Consider the top-right element. If it's smaller than target, then the first row is eliminated. If it's larger than the target, then the last column is eliminated.

    Runtime: 280ms.

     1 class Solution {
     2 public:
     3     bool searchMatrix(vector<vector<int>>& matrix, int target) {
     4         if(matrix.size() == 0 ) return false;
     5         
     6         int m = matrix.size(), n = matrix[0].size();
     7         int row = 0, col = n - 1;
     8         
     9         while(row < m && col >= 0){
    10             if(matrix[row][col] < target) row++;
    11             else if(matrix[row][col] > target) col--;
    12             else return true;
    13         }
    14         return false;
    15     }
    16 };
  • 相关阅读:
    scnner02 (nextLine)
    Scanner01
    Spring 框架 (初学)
    查询自己写了多少行代码
    jdbc事务
    jdbc(预编译插入数据)
    jdbc(java连接数据库)
    监听器扩展
    listener(监听器)
    Filter过滤器
  • 原文地址:https://www.cnblogs.com/amazingzoe/p/4672912.html
Copyright © 2011-2022 走看看