leetcode [74] 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.

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

题目大意：

矩阵的值是按行进行递增的，每一行从左到右也是递增的，查找一个target值是否在矩阵中存在。

 

解法：

我一想到这个题目，就想到使用二分查找，但是这是一个矩阵，并不是一个数组，所以将矩阵变成数组就可以了啊。

C++：

class Solution {
public:
    bool searchVector(vector<int>v,int target,int start,int end){
        if(end<start) return false;
        int mid=(end-start)/2+start;
        if(v[mid]==target) return true;
        else if(v[mid]<target) return searchVector(v,target,mid+1,end);
        else return searchVector(v,target,start,mid-1);
    }

    bool searchMatrix(vector<vector<int>>& matrix, int target) {
        if(matrix.empty()||matrix[0].empty()) return false;
        vector<int>v;
        for(int i=0;i<matrix[0].size();i++) v.push_back(matrix[0][i]);
        for(int i=1;i<matrix.size();i++){
            v.insert(v.end(),matrix[i].begin(),matrix[i].end());
        }
        return searchVector(v,target,0,v.size()-1);
    }
};

这种二分查找虽然是对的，但我总觉得，这样做不太好，用了很多额外的空间，应该转换成为矩阵的二分查找。

改进后的C++：

class Solution {
public:
    bool searchMatrix(vector<vector<int>>& matrix, int target) {
        if(matrix.empty()||matrix[0].empty()) return false;
        int m=matrix.size();
        int n=matrix[0].size();

        if (matrix[0][0]>target||matrix[m-1][n-1]<target) return false;
        int start=0,end=m*n-1;
        int mid,midRow,midCol;
        while(start<=end){
            mid=start+(end-start)/2;
            midCol=mid%n;
            midRow=mid/n;
            if(matrix[midRow][midCol]<target) start=mid+1;
            else if(matrix[midRow][midCol]>target) end=mid-1;
            else return true;
        }
        return false;
    }
};

Python:

class Solution(object):
    def searchMatrix(self, matrix, target):
        """
        :type matrix: List[List[int]]
        :type target: int
        :rtype: bool
        """
        if not matrix or target is None:
            return False
        m,n=len(matrix),len(matrix[0])
        mid=midCol=midRow=0
        start,end=0,m*n-1
        while start<=end:
            mid=(start+end)/2
            midCol=mid%n
            midRow=mid/n
            if matrix[midRow][midCol]<target:
                start=mid+1
            elif matrix[midRow][midCol]>target:
                end=mid-1
            else:
                return True
        return False