[leetcode 74] Search a 2D Matrix

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本文介绍了一种高效的算法，用于在一个m x n的二维有序矩阵中查找特定数值。该矩阵的特点是每一行从左到右元素递增排序，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C++实现代码。

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 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.

class Solution {
public:
    bool searchMatrix(vector<vector<int> > &matrix, int target) {
        const int m = matrix.size();
        const int n = matrix[0].size();
        int l = 0;
        int r = m*n;
        while (l < r) {
            int mid = l +(r-l)/2;
            int val = matrix[mid/n][mid%n];
            if (val == target) {
                return true;
            } else if (val > target) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return false;
    }
};