LeetCode 240. Search a 2D Matrix II

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.

m x n矩阵中，每一行和每一列都是从小到大排列，要求判断一个值是否属于该矩阵。

解决思路：从右上角或从左下角的元素开始搜索。以右上角为例：

如果元素的值等于target，则返回true；

如果元素的值小于target，则应该搜索左边的元素；

如果元素的值大于target,则应当搜索下面的元素；

当左边或者下面没有元素的时候，结束循环，返回false。

代码实现如下：

class Solution {public:bool searchMatrix(vector< vector<int> >& matrix, int target) {if(matrix.size()==0 || matrix[0].size()==0) return false;  int m=0, n=matrix[0].size()-1;  while(m<matrix.size() && n>=0) {  int x = matrix[m][n];  if(target == x) return true;  else if(target < x) --n;  else ++m;  }  return false;  }};