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【leetcode刷题笔记】Search a 2D Matrix

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.

题解：每次用右上角的元素跟target比较，有三种情况：

相等，说明找到了target；
右上角元素比target元素大，那么说明target在第一行，递归的搜索第一行。
右上角元素比target元素小，那么说明target在第二行及以后，递归的搜索如下图所示区域：

代码如下：

 1 public class Solution {
 2     private boolean IfFind = false;
 3     private void RightCornerRecur(int[][] matrix,int target,int m_start,int m_end,int n_start,int n_end){
 4         if(m_start > m_end || n_start > n_end)
 5             return;
 6             
 7         if(m_start < 0 || n_start < 0 || m_end >= matrix.length || n_end >= matrix[0].length)
 8             return;
 9         
10         if(matrix[m_start][n_end] == target){
11             IfFind = true;
12             return;
13         }
14         if(matrix[m_start][n_end] > target)
15             RightCornerRecur(matrix, target, m_start, m_start, n_start, n_end-1);
16         else {
17             RightCornerRecur(matrix, target, m_start+1,m_end, n_start, n_end);
18         }
19 
20     }
21     public boolean searchMatrix(int[][] matrix, int target) {
22         RightCornerRecur(matrix, target,0,matrix.length-1, 0, matrix[0].length-1);
23         return IfFind;
24     }
25 }

右上角可以用来比较剪掉一些元素，左下角同样可以，下面的代码中加上了左下角元素与target元素的比较，最终的运行时间是384ms，而上述只考虑右上角的运行时间是472ms。

 1 public class Solution {
 2     private boolean IfFind = false;
 3     private void RightCornerRecur(int[][] matrix,int target,int m_start,int m_end,int n_start,int n_end){
 4         
 5         if(m_start > m_end || n_start > n_end)
 6             return;
 7             
 8         if(m_start < 0 || n_start < 0 || m_end >= matrix.length || n_end >= matrix[0].length)
 9             return;
10         
11     
12         if(matrix[m_start][n_end] == target){
13             IfFind = true;
14             return;
15         }
16         
17         if(matrix[m_start][n_end] > target)
18             RightCornerRecur(matrix, target, m_start, m_start, n_start, n_end-1);
19         
20         else {
21             if(matrix[m_end][0] == target){
22                 IfFind = true;
23                 return;
24             }
25             if(matrix[m_end][0] < target)
26                 RightCornerRecur(matrix, target, m_end, m_end, n_start+1, n_end);
27             else
28                 RightCornerRecur(matrix, target, m_start+1,m_end-1, n_start, n_end);
29         }
30 
31     }
32     public boolean searchMatrix(int[][] matrix, int target) {
33         RightCornerRecur(matrix, target,0,matrix.length-1, 0, matrix[0].length-1);
34         return IfFind;
35     }
36 }