原题:
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
.
解决方法:
由题意可知,每一行或每一列是有序的,可以采用十字交叉搜索的方法,从右上角的点开始:
- 如果相等,则可以直接返回;
- 如果该点的值小于目标值,则表明在前面的列,j--。
-如果该点的值大于目标值,则表明在后面的行,i++。
代码:
bool searchMatrix(vector<vector<int>>& matrix, int target) {
int row = matrix.size();
if (row < 1)
return false;
int col = matrix[0].size();
if (col < 1)
return false;
int i = 0, j = col -1;
while( i < row && j >= 0){
if (matrix[i][j] == target)
return true;
else if (matrix[i][j] < target)
++i;
else
--j;
}
return false;
}