118：Search a 2D Matrix

最新推荐文章于 2020-03-13 10:48:44 发布

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本文介绍了一种高效的矩阵搜索算法，该算法能在排序的二维矩阵中快速查找目标值。矩阵的每一行从左到右递增排序，并且每行的第一个元素大于前一行的最后一个元素。文章提供了一个具体的例子和实现代码，采用二分查找法，时间复杂度为O(log(nm))，空间复杂度为O(1)。

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题目：Write an efficient algorithm that searches for a value in an mn 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.

可以用二分查找解决，代码如下：

// 时间复杂度 O(log(nm))，空间复杂度 O(1)
class Solution {
public:
        bool searchMatrix(vector<vector<int>>& matrix, int target) {
                if (matrix.empty()) return false;
                const size_t m = matrix.size();
                const size_t n = matrix[0].size();

                int first = 0;
                int last = m * n;
                while (first != last) {
                        int mid = first + (last - first) / 2;
                        int value = matrix[mid / n][mid % n];

                        if (value == target)
                                return true;
                        else if (value < target)
                                first = mid + 1;
                        else
                                last = mid;
                }
                return false;
        }
};