[Leetcode] 240. Search a 2D Matrix II 解题报告

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.

思路：

由于行和列都是有序的，所以我们可以从二维数组的右上角开始搜索。如果值刚好等于target，则返回；如果值小于target，则往下移动；否则往左移动。假设二维数组的行个数和列个数分别是m和n，则时间复杂度是O(m+n)。空间复杂度是O(1)。

代码：

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

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