LeetCode74 Search a 2D Matrix

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. （Medium）

分析：

考察题目给的二维数组的特点发现，其本质跟一个排好序的一维数组没有区别，所以就是一个二分查找问题。

对于mid，其对应的点是matrix[mid / n][mid % n]

代码：

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