Search a 2D Matrix <leetcode>

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.

思路:首先遍历每一行的首元素,找到目标如果存在,应该在哪一行,然后在行里进行搜索,本来在这里想用二分查找,但是直接扫了一遍也ac了,代码如下(没有进行优化):

 1 class Solution {
 2 public:
 3     bool searchMatrix(vector<vector<int> > &matrix, int target) {
 4         int index=0;
 5         if(matrix.size()==0)
 6         {
 7             return false;
 8         }
 9         if(matrix.size()==1)
10         {
11             for(int i=0;i<matrix[0].size();i++)
12             {
13                 if(target==matrix[0][i])
14                 {
15                     return true;
16                 }
17             }
18             return false;
19         }
20         for(int i=0;i<matrix.size();i++)
21         {
22             if(target>matrix[i][0])
23             {
24                 index++;
25             }
26             else if(target==matrix[i][0])
27             {
28                 return true;
29             }
30         }
31         if(index==0)  return false;
32         for(int i=0;i<matrix[0].size();i++)
33         {
34             if(target==matrix[index-1][i])
35             {
36                 return true;
37             }
38         }
39         return false;
40         
41     }
42 };
原文地址:https://www.cnblogs.com/sqxw/p/3960185.html