[LeetCode] Search a 2D Matrix

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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:

• 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;
    }
};

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[LeetCode] Search a 2D Matrix

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原文地址：http://www.cnblogs.com/Xylophone/p/3796758.html