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

.
给一个在两个维度上都递增的矩阵，判断矩阵中是否存在一个数。
如果用二分需要遍历每一行，对每一行二分判断target是否存在，复杂度为O(n*log(m))。

如果从矩阵的右上角或者左下角开始，则相邻的两个方向可以构成一个决策集，一个方向一定比当前数小，一个方向一定比当前数大，用这种方法复杂度为O(m + n).

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