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.
1 class Solution { 2 public: 3 bool searchMatrix(vector<vector<int> >& matrix, int target) { 4 int col=matrix[0].size(); 5 int row=matrix.size(); 6 7 int targetrow=0; 8 int flag=0; 9 10 11 if(0==col-1&&0==row-1) 12 { 13 if(target==matrix[0][0]) 14 return true; 15 else 16 return false; 17 } 18 19 20 for(int i=0;i<row;i++) 21 { 22 if(target<matrix[0][0]) 23 { 24 return false; 25 } 26 if(target==matrix[i][0]) 27 { 28 return true; 29 } 30 if(target<matrix[i][0]&&i>=1) 31 { 32 flag=1; 33 targetrow=i-1; 34 break; 35 } 36 if(target<matrix[row-1][col-1]) 37 { 38 flag=1; 39 targetrow=row-1; 40 } 41 if(target==matrix[row-1][col-1]) 42 { 43 return true; 44 } 45 flag=0; 46 } 47 48 49 for(int i=0;i<col;i++) 50 { 51 if(target==matrix[targetrow][i]) 52 { 53 flag=1; 54 break; 55 } 56 flag=0; 57 } 58 59 if(0==flag) 60 return false; 61 else 62 return true; 63 } 64 };
原文：http://www.cnblogs.com/jawiezhu/p/4474682.html