题目

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.

解题思路

分治法求解，首先根据矩阵的特点，横向和纵向是分别升序的。可以从右上角或者左下角进行判断。拿左下角来说
1. 如果比target大，那么需要往上走，因为当前行的右边都比它大
2. 如果比target小，那么需要往右走，因为当前列的上面比它都小
3. 相等则命中

每次都是通过减少一行或者一列来向前搜索，复杂度就是O（M+N）M,N为矩阵行列数

代码

class Solution {
public:
    bool searchMatrix(vector<vector<int>>& matrix, int target) {

         int row=matrix.size();
         if(row == 0)  return false;

         int column=matrix[0].size();
         int j=0;
         int i=row-1;

         while(j<column&&i>=0)
         {
             int temp=matrix[i][j];
             if(temp==target)
             {
                 return true;
             }
             if(temp>target)
             {
                 i--;
             }
             else
             {
                 j++;
             }
         }
         return false;
    }
};