LeetCode240:Search a 2D Matrix II

最新推荐文章于 2022-11-16 19:19:52 发布

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本文介绍了一个高效的算法，用于在一个特殊矩阵中查找特定值。该矩阵的每一行从左到右递增排序，每一列从上到下递增排序。文章提供了一个时间复杂度为O(m+n)的解决方案。

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.

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//时间复杂度为O(m+n)
class Solution{
public:
	bool searchMatrix(vector<vector<int>>& matrix, int target) {
		int m = matrix.size();
		int n = matrix[0].size();

		int i = 0, j = n - 1;
		while (i < m && j >= 0)
		{
			if (matrix[i][j] == target)
				return true;
			else if (matrix[i][j] > target)
				j--;
			else
				i++;
		}
		return false;
	}
};