【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：

1. 相等，说明找到了target；
2. 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。
3. 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：

 public class Solution {
     private boolean IfFind = false;
     private void RightCornerRecur(int[][] matrix,int target,int m_start,int m_end,int n_start,int n_end){
         if(m_start > m_end || n_start > n_end)
             return;

         if(m_start < 0 || n_start < 0 || m_end >= matrix.length || n_end >= matrix[0].length)
             return;

         if(matrix[m_start][n_end] == target){
             IfFind = true;
             return;
         }
         if(matrix[m_start][n_end] > target)
             RightCornerRecur(matrix, target, m_start, m_start, n_start, n_end-1);
         else {
             RightCornerRecur(matrix, target, m_start+1,m_end, n_start, n_end);
         }

     }
     public boolean searchMatrix(int[][] matrix, int target) {
         RightCornerRecur(matrix, target,0,matrix.length-1, 0, matrix[0].length-1);
         return IfFind;
     }
 }

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。

 public class Solution {
     private boolean IfFind = false;
     private void RightCornerRecur(int[][] matrix,int target,int m_start,int m_end,int n_start,int n_end){

         if(m_start > m_end || n_start > n_end)
             return;

         if(m_start < 0 || n_start < 0 || m_end >= matrix.length || n_end >= matrix[0].length)
             return;

         if(matrix[m_start][n_end] == target){
             IfFind = true;
             return;
         }

         if(matrix[m_start][n_end] > target)
             RightCornerRecur(matrix, target, m_start, m_start, n_start, n_end-1);

         else {
             if(matrix[m_end][0] == target){
                 IfFind = true;
                 return;
             }
             if(matrix[m_end][0] < target)
                 RightCornerRecur(matrix, target, m_end, m_end, n_start+1, n_end);
             else
                 RightCornerRecur(matrix, target, m_start+1,m_end-1, n_start, n_end);
         }

     }
     public boolean searchMatrix(int[][] matrix, int target) {
         RightCornerRecur(matrix, target,0,matrix.length-1, 0, matrix[0].length-1);
         return IfFind;
     }
 }