搜索循环排序矩阵和运行复杂度

Question

该方法给出 NxN 矩阵总是 2 的幂和一个数字，如果找到 num 则它将 return 为真例如 4x4 大小：

这是我写的：

public class Search {
public static boolean Search (int [][] matrix, int num) 
{
int value = matrix.length / 2;
int first_quarter_pivot = matrix[value-1][0]; // represents highest number in first quarter
int second_quarter_pivot = matrix[value-1][value]; // represents highest number in second quarter
int third_quarter_pivot = matrix[matrix.length-1][value]; // represents highest number in third quarter
int fourth_quarter_pivot = matrix[matrix.length-1][0]; // represents highest number in fourth quarter
boolean isBoolean = false; 
int i=0;
int j;  



// if the num is not in the range of biggest smallest number it means he can`t be there.    
 if(!(num >= first_quarter_pivot) && (num <= fourth_quarter_pivot)) {
 return false;
}
// if num is one of the pivots return true;
if((num == first_quarter_pivot || (num ==second_quarter_pivot)) 
|| (num == third_quarter_pivot) || (num == fourth_quarter_pivot ))
return true;

// if num is smaller than first pivot it means num is the first quarter,we limit the search to first quarter.
// if not smaller move to the next quarter pivot
if(num < first_quarter_pivot){{
    j =0;
    do
   
        if(matrix[i][j] == num) {
               isBoolean = true;
               break;
               
               
            }
           else if((j == value)) {
                        j = 0;
                        i++;
                    }
                        else if(matrix[i][j] != num){
                         j++;
                        }
                        while(isBoolean != true) ;
                    }
      
              return isBoolean;

            }

    
// if num is smaller than second pivot it means num is the second quarter,we limit the search to second quarter.
// if not smaller move to the next quarter pivot    
if(num < second_quarter_pivot){{
    j = value;// start (0,value) j++ till j=value
    do
    
        if(matrix[i][j] == num) {
               isBoolean = true;
               break;
            }
           else if((j == matrix.length-1)) {
                        j = value;
                        i++;
                    }
                        else if(matrix[i][j] != num){
                         j++;
                        }
                        while(isBoolean != true) ;
               
}
return isBoolean;
}
            
            // if num is smaller than third pivot it means num is the third quarter,we limit the search to third quarter.
// if not smaller move to the next quarter pivot
            
if(num < third_quarter_pivot){{
i = value;              
j = value;// start (0,value) j++ till j=value
    do
    
        if(matrix[i][j] == num) {
               isBoolean = true;
               break;
            }
           else if((j == matrix.length-1)) {
                        j = value;
                        i++;
                    }
                        else if(matrix[i][j] != num){
                         j++;
                        }
                        while(isBoolean != true) ;
           
            }
 return isBoolean;
      
  }     
          // if num is smaller than fourth pivot it means num is the fourth quarter,we limit the search to fourth quarter.
// number must be here because we verfied his existence in the start.
  if(num < fourth_quarter_pivot){
    i = value;              
    j = 0;// start (0,value) j++ till j=value
    do
    
        if(matrix[i][j] == num) {
               isBoolean = true;
               break;
            }
           else if((j == value)) {
                        j = 0;
                        i++;
                    }
                        else if(matrix[i][j] != num){
                         j++;
                        }
                        while(isBoolean != true) ;
 
} 
return isBoolean;    

}

}

我尝试做的事情：找到想要的号码在哪个季度，然后检查通过移动 j++ 直到它达到极限，比 i++ 相同的季度直到找到随着每个季度的限制变化，我可以t understand if run time complexity is O(n^2) or lower? and will it be better do create one dimensional array and and move on the quarter this way: move right until limit,one down,move left until limit and i我有一个排序数组和二进制搜索

Answer 1

如果可以将数组映射到矩阵，则可以使用普通的二分查找。

您可以像这样定义翻译 table 来实现：

X = [0, 0, 1, 1, 0, 0, 1, 1, 2, 2, 3, 3, 2, 2, 3, 3, ...]
Y = [0, 1, 1, 0, 2, 3, 3, 2, 2, 3, 3, 2, 0, 1, 1, 0, ...]

最终程序如下所示。

static final int MAX_N = 64;
static final int MAX_NN = MAX_N * MAX_N;
static final int[] DX = {0, 0, 1, 1};
static final int[] DY = {0, 1, 1, 0};
static final int[] X = new int[MAX_NN];
static final int[] Y = new int[MAX_NN];

static {  // initialize X and Y
    for (int i = 0; i < MAX_NN; ++i) {
        int x = 0, y = 0;
        for (int t = i, f = 0; t > 0; ++f) {
            int mod = t & 3;
            x += DX[mod] << f; y += DY[mod] << f;
            t >>= 2;
        }
        X[i] = x; Y[i] = y;
    }
}

public static boolean Search(int [][] matrix, int num) {
    int n = matrix.length, nn = n * n;
    int lower = 0;
    int upper = nn - 1;
    while (lower <= upper) {
        int mid = (lower + upper) / 2;
        int value = matrix[X[mid]][Y[mid]];
        if (value == num)
            return true;
        else if (value < num)
            lower = mid + 1;
        else
            upper = mid - 1;
    }
    return false;
}

和

public static void main(String[] args) {
    int[][] matrix = {
        {1, 3, 7, 9},
        {6, 4, 15, 11},
        {36, 50, 21, 22},
        {60, 55, 30, 26},
    };
    // case: exists
    System.out.println(Search(matrix, 1));
    System.out.println(Search(matrix, 60));
    System.out.println(Search(matrix, 11));
    // case: not exists
    System.out.println(Search(matrix, 0));
    System.out.println(Search(matrix, 70));
    System.out.println(Search(matrix, 20));
}

输出：

true
true
true
false
false
false

搜索循环排序矩阵和运行复杂度

Search in circular sorted martix and run complexity

java

arrays

performance

runtime

matrix

搜索循环排序矩阵和 运行 复杂度

Search in circular sorted martix and run complexity

java

arrays

performance

runtime

matrix

搜索循环排序矩阵和运行复杂度