leetcode_c++:Divide and Conquer： . Search a 2D Matrix II（240）

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于 2016-08-28 15:35:10 发布

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本文介绍了一种高效的矩阵搜索算法，该算法能在排序的二维矩阵中查找特定的目标值。矩阵的每一行从左到右递增排序，每一列从上到下递增排序。通过示例演示了如何使用该算法进行搜索，并提供了具体的实现代码。

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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 in ascending from left to right.
Integers in each column are sorted in ascending from top to bottom.
For example,

Consider the following matrix:

[
  [1,   4,  7, 11, 15],
  [2,   5,  8, 12, 19],
  [3,   6,  9, 16, 22],
  [10, 13, 14, 17, 24],
  [18, 21, 23, 26, 30]
]

Given target = 5, return true.

Given target = 20, return false.

class Solution {  
public:  
    /** 
     * @param matrix: A list of lists of integers 
     * @param target: An integer you want to search in matrix 
     * @return: An integer indicate the total occurrence of target in the given matrix 
     */  
    int searchMatrix(vector<vector<int> > &matrix, int target) {  
        // write your code here  

        if (matrix.size() == 0)  
            return 0;  

        int count = 0;  
        int rows = matrix.size();  
        int colums = matrix[0].size();  

        int curRowPos = 0;  
        int curColumPos = colums-1;  

        while (curRowPos < rows && curColumPos >= 0)  
        {  
            if (matrix[curRowPos][curColumPos] > target)  
            {  
                curColumPos--;  
            }  
            else if (matrix[curRowPos][curColumPos] < target)  
            {  
                curRowPos++;  
            }  
            else  
            {  
                count++;  
                curColumPos--;  
            }  
        }  

        return count;  
    }  
};