Search a 2D Matrix

最新推荐文章于 2022-08-05 12:18:43 发布

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最新推荐文章于 2022-08-05 12:18:43 发布

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原文链接：http://www.cnblogs.com/sankexin/p/5865957.html

本文介绍了一种高效的矩阵搜索算法，该算法能在排序的二维矩阵中快速查找目标值。矩阵的每一行从左到右递增排序，并且每行的第一个元素大于前一行的最后一个元素。文章通过示例展示了如何实现这一算法。

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 from left to right.
The first integer of each row is greater than the last integer of the previous row.

For example,

Consider the following matrix:

[
  [1,   3,  5,  7],
  [10, 11, 16, 20],
  [23, 30, 34, 50]
]

Given target = 3, return true.

 1 class Solution {
 2 public:
 3     bool searchMatrix(vector<vector<int>>& matrix, int target) {
 4         if(matrix.size() == 0){
 5             return false;
 6         }
 7         if(matrix[0].size() == 0){
 8             return false;
 9         }
10         
11         int rowNumber = 0;
12         int colNumber = matrix[0].size()-1;
13         
14         while(rowNumber < matrix.size() && colNumber >= 0){
15             
16             if(matrix[rowNumber][colNumber] > target){
17                 colNumber--;
18             }else if(matrix[rowNumber][colNumber] < target){
19                 rowNumber++;
20             }else{
21                 return true;
22             }
23         }
24         return false;
25     }
26 };