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  • [LeetCode] Search a 2D Matrix

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