LeetCode----240. Search a 2D Matrix II (M)

1 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

.

2 分析

对一个有序的矩阵进行查找，有点类似二分法的思想，矩阵的第一个数和最后一个数分别是最小和最大的数

选择从对角线的元素开始查找即可，例如从左下角的元素开始查找，如果target > matrix[i][j]，则j++；否则i--

3 C++实现

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

if(matrix.empty() || matrix[0].empty())
return false;

int m = matrix.size();
int n = matrix[0].size();

if(target > matrix[m - 1][n - 1] || target < matrix[0][0])
return false;

//从左下角的数开始找
int i = m - 1;
int j = 0;

while(i >= 0 && j < n) {
if(target == matrix[i][j])
return true;
else if(target < matrix[i][j])
i--;
else j++;

}

return false;

}
};