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Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:
For example,
Consider the following matrix:
[ [1, 3, 5, 7], [10, 11, 16, 20], [23, 30, 34, 50] ]
Given target = 3, return true.
分析：将二维数组看为一个一维sorted数组，然后用bianry search。时间复杂度为O(log(m*n)),代码如下：
1 class Solution { 2 public: 3 bool searchMatrix(vector<vector<int> > &matrix, int target) { 4 int m = matrix.size(); 5 if(m == 0) return false; 6 int n = matrix[0].size(); 7 if(n == 0) return false; 8 9 int l = 0, r = m * n - 1; 10 while(l <= r){ 11 int mid = (l + r)/2; 12 int row = mid/n, col = mid%n; 13 if(matrix[row][col] == target) return true; 14 if(matrix[row][col] < target) l = mid + 1; 15 else r = mid - 1; 16 } 17 18 return false; 19 } 20 };
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原文地址：http://www.cnblogs.com/Kai-Xing/p/4206453.html
