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.

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首先二分查找每行的第一个元素,确定了行号后,再在当前行内进行二分查找

值得一提的是其时间复杂度为 O(log(m) + log(n)) = O(log(m*n))

bool searchMatrix(vector<vector<int>>& matrix, int target) {
int beg = , mid, end = matrix.size()-;
while (beg <= end)
{
mid = (beg + end) >> ;
if (target < matrix[mid][])
end = mid - ;
else if (target > matrix[mid][])
beg = mid + ;
else
return true;
}
int row = end;
if (row < )
return false;
beg = ;
end = matrix[row].size();
while (beg <= end)
{
mid = (beg + end) >> ;
if (target < matrix[row][mid])
end = mid - ;
else if (target > matrix[row][mid])
beg = mid + ;
else
return true;
}
return false;
}

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