medium 剑指 Offer 04. 二维数组中的查找 遍历 线性查找

class Solution {
public:
    bool findNumberIn2DArray(vector>& matrix, int target) {
        if( matrix.size()==0 || matrix.size()==0 ) {
            return false;
        }
        int rows = matrix.size();  // 不能一开始就赋值，rows,colunms,要先判断非空矩阵，否则赋值出错
        int columns =matrix[0].size();
        for(int i=0; i 
python 
class Solution:
    def findNumberIn2DArray(self, matrix: List[List[int]], target: int) -> bool:
        if len(matrix)==0 or len(matrix[0])==0:
            return False
        rows = len(matrix)
        columns = len(matrix[0])
        for i in range(0, rows):
            for j in range(0, columns):
                if matrix[i][j] == target:
                    return True
                    
        return False

 
 
 
线性查找： 
 
c++ 
class Solution {
public:
    bool findNumberIn2DArray(vector>& matrix, int target) {
        if(matrix.size()==0||matrix[0].size()==0){
            return false;
        }
        int rows = matrix.size();  // 不能一开始就赋值，rows,colunms,要先判断非空矩阵，否则赋值出错
        int columns =matrix[0].size();

        // 从右上角开始遍历
        int i = 0;
        int j = columns-1;
        while( i=0){
            if(matrix[i][j] == target){
                return true;
            }else if( matrix[i][j] > target ){
                j--;
            }else{  // matrix[i][j] < target 
                i++;
            }
        }

        return false;
    }
};

 
python 
class Solution:
    def findNumberIn2DArray(self, matrix: List[List[int]], target: int) -> bool:
        if len(matrix)==0 or len(matrix[0])==0:
            return False

        rows = len(matrix)
        columns = len(matrix[0])
        i = 0
        j = columns-1
        while i=0:
            if matrix[i][j] == target:
                return True
            elif matrix[i][j] > target:
                j -= 1
            else:
                i += 1

        return Fals