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
.
编程思想:从矩阵的右上角的元素开始与目标元素比较大小,矩阵右上角元素,比目标元素大的,删除矩阵右上角元素所在列,否则删除所在行,直到找到目标元素或者删除所有行列。
class Solution { public: bool searchMatrix(vector<vector<int> > &matrix, int target) { if(matrix.empty()) return false; int rows = matrix.size(); int columns = matrix[0].size(); int row = 0,column = columns-1; while(row<rows && column>=0){ if(matrix[row][column]==target) return true; else if(matrix[row][column]>target) column--; else row++; }//end while return false; } };