Search a 2D Matrix

最新推荐文章于 2025-01-17 23:23:30 发布

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最新推荐文章于 2025-01-17 23:23:30 发布

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原文链接：http://www.cnblogs.com/tianzeng/p/10865176.html

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.

Example 1:
Input:
matrix = [
  [1,   3,  5,  7],
  [10, 11, 16, 20],
  [23, 30, 34, 50]
]
target = 3
Output: true


Example 2:
Input:
matrix = [
  [1,   3,  5,  7],
  [10, 11, 16, 20],
  [23, 30, 34, 50]
]
target = 13
Output: false

code

class Solution
{
public:
    bool searchMatrix(vector<vector<int>>& matrix, int target)
    {
        if(matrix.empty()||matrix.at(0).empty()||target<matrix.at(0).at(0)||target>matrix.at(matrix.size()-1).at(matrix.at(0).size()-1))
            return false;

        int low=0,height=matrix.size()-1;
        int mid=0;
        while(low<=height)
        {
            mid=(low+height)/2;
            if(target>=matrix.at(mid).at(0))
            {
                if(mid==matrix.size()-1||(mid<matrix.size()-1&&target<matrix.at(mid+1).at(0)))
                    break;
            }

            if(target<matrix.at(mid).at(0))
                height=mid-1;
            else if(target>matrix.at(mid).at(0))
                low=mid+1;
        }

        int pos=mid;
        low=0;
        height=matrix.at(mid).size()-1;
        while(low<=height)
        {
            mid=(low+height)/2;
            if(matrix.at(pos).at(mid)==target)
                return true;

            if(matrix.at(pos).at(mid)>target)
                height=mid-1;
            else if(matrix.at(pos).at(mid)<target)
                low=mid+1;
        }
        return false;
    }
};

code2

class Solution
{
public:
    bool searchMatrix(vector<vector<int>>& matrix, int target)
    {
        if(matrix.empty()||matrix.at(0).empty()||target<matrix.at(0).at(0)||target>matrix.at(matrix.size()-1).at(matrix.at(0).size()-1))
            return false;

        int row=matrix.size(),col=matrix.at(0).size();
        int left=0,right=row*col-1;
        while(left<=right)
        {
            int mid=(left+right)/2;
            int i=mid/col;
            int j=mid%col;
            if(matrix.at(i).at(j)==target)
                return true;
            else if(matrix.at(i).at(j)<target)
                left=mid+1;
            else if(matrix.at(i).at(j)>target)
                right=mid-1;
        }
        return false;
    }
};