Search a 2D Matrix

来源：互联网发布：碧海潮生曲知乎编辑：程序博客网时间：2024/05/18 17:03

原题如下：
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.

这个题其实还是很简单的，直接暴力也能AC， C++代码如下：

    bool searchMatrix(vector<vector<int>>& matrix, int target) {        if(matrix.size() < 1 || matrix[0].size() < 1)             return false;        int m = matrix.size(), n = matrix[0].size();        for(int i=0; i<m; i++){            for(int j=0; j<n; j++){                if(matrix[i][j] == target){                    return true;                }            }        }        return false;    }

不过既然有规律可循，那肯定要找到一种高效的方法了，可以考虑把矩阵“展平”，这不就是二分查找嘛。这样一下子就可以把时间从O(MN)降到了O(Log(MN))了。AC过的C++代码如下：

    bool searchMatrix(vector<vector<int>>& matrix, int target) {        if(matrix.size() < 1 || matrix[0].size() < 1)             return false;        int m = matrix.size(), n = matrix[0].size();        int left = 0, right = m * n - 1;        while(left <= right){            int mid = (left + right) / 2;            int i = mid /n, j = mid % n;            if(matrix[i][j] == target)               return true;            else if(matrix[i][j] > target){                right = mid - 1;            }else{                left = mid + 1;            }        }        return false;    }

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