题目：
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.
思路：二分查找，先找到行数再找列数。
代码：
1 class Solution { 2 public: 3 bool searchMatrix(vector<vector<int> > &matrix, int target) { 4 // Start typing your C/C++ solution below 5 // DO NOT write int main() function 6 int m = matrix.size(); 7 int n = matrix[0].size(); 8 if (matrix[0][0]>target || matrix[m-1][n-1]<target) return false; 9 int top = 0; 10 int down = m-1; 11 int mid; 12 while (top <= down) { 13 mid = (top+down)/2; 14 if (target == matrix[mid][0]) return true; 15 if (target > matrix[mid][0]) top = mid+1; 16 else down = mid-1; 17 } 18 int row = down; 19 int left = 0; 20 int right = n-1; 21 while (left <= right) { 22 mid = (left+right)/2; 23 if (target == matrix[row][mid]) return true; 24 if (target > matrix[row][mid]) left = mid+1; 25 else right = mid-1; 26 } 27 return false; 28 } 29 };
转载于:https://www.cnblogs.com/tanghulu321/archive/2013/05/03/3055787.html
