如何從二維數組中找到索引

Question

我想要做的是打印一個二維數組中的最大數字，它是索引位置。我能找到最大的數字，但我似乎無法弄清楚如何打印它的索引位置。無論如何，這是我到目前爲止：

public static void main(String[] args) { 
    int[][] arr = {{4, 44, 5, 7, 63, 1}, {7, 88, 31, 95, 9, 6}, {88, 99, 6, 5, 77, 4}}; 

    double max = arr[0][0]; 
    for (int i = 0; i < arr.length; i++) { 
     for (int j = 0; j < arr.length; j++) { 
      if (arr[i][j] > max) { 
       max = arr[i][j]; 

      } 
     } 
    } 
    System.out.println(max); 
    System.out.println(i + j); //No idea what I should be doing here, just trying out everything I can think of 

Answer 1

存儲我，j當你更新最大。

Answer 2

你有一個二維數組，因此你需要知道兩個索引。把它們加在一起不會做，因爲你失去了哪一個。這個怎麼樣：

System.out.println("[" + i + "][" + j + "]"); 

Answer 3

現在，你應該始終得到2 * arr.length作爲最終值。這不是你可能在尋找的東西。它看起來像你想知道最大值的座標。要做到這一點，你需要緩存率的值，然後在以後使用它們：

public static void main(String[] args) { 
    int[][] arr = {{4, 44, 5, 7, 63, 1}, {7, 88, 31, 95, 9, 6}, {88, 99, 6, 5, 77, 4}}; 
    int tmpI = 0; 
    int tmpJ = 0; 
    double max = arr[0][0]; 
    // there are some changes here. in addition to the caching 
    for (int i = 0; i < arr.length; i++) { 
     int[] inner = arr[i]; 
     // caches inner variable so that it does not have to be looked up 
     // as often, and it also tests based on the inner loop's length in 
     // case the inner loop has a different length from the outer loop. 
     for (int j = 0; j < inner.length; j++) { 
      if (inner[j] > max) { 
       max = inner[j]; 
       // store the coordinates of max 
       tmpI = i; tmpJ = j; 
      } 
     } 
    } 
    System.out.println(max); 
    // convert to string before outputting: 
    System.out.println("The (x,y) is: ("+tmpI+","+tmpJ+")"); 

Answer 4

這將是，如果你想要一個索引到一個扁平化陣列：

public static void main (String[] args) throws java.lang.Exception 
{ 
     int[][] arr = {{4, 44, 5, 7, 63, 1}, {7, 88, 31, 95, 9, 6}, {88, 99, 6, 5, 77, 4}}; 
      int[] flattened = new int[6*3]; // based off above 
      int maxIndex = 0; 
      double max = arr[0][0]; 
      for (int i = 0; i < arr.length; i++) { 
       for (int j = 0; j < arr.length; j++) { 
        flattened[i + j] = arr[i][j]; 
        if (arr[i][j] > max) { 
         max = arr[i][j]; 
         maxIndex = i+j; 
        } 
       } 
     } 
    System.out.println(max); 
    System.out.println(flattened [maxIndex]); 
} 

Answer 5

唐不確定你是否實現了有效的算法，但是當你設置最大值時，爲什麼你只是不把索引i，j保存在另一個變量中。 這很簡單。

if (arr[i][j] > max) { 
    max = arr[i][j]; 
    maxX = i; 
    maxY = j; 
} 

供參考如果你想看看「插入排序」算法，如果你想更好的實施。

Answer 6

小心你的數組尺寸！你們大多數人的第二個陳述是錯誤的。它應該去達改編[I]。長度：

for (int i = 0; i < arr.length; i++) { 
    for (int j = 0; j < arr[i].length; j++) { 
     if (arr[i][j] > max) { 
      max = arr[i][j]; 
      tmpI = i; tmpJ = j; 
     } 
    } 
} 

Answer 7

int[][] arr = {{4, 44, 5, 7, 63, 1}, {7, 88, 31, 95, 9, 6}, {88, 99, 6, 5, 77, 4}}; 

int max = arr[0][0]; 
int maxI = 0, maxJ = 0; 

for (int i = 0; i < arr.length; i++) { 
    for (int j = 0; j < arr.length; j++) { 
     if (arr[i][j] > max) { 
      max = arr[i][j]; 
      maxI = i; 
      maxJ = j; 
     } 
    } 
} 
System.out.println(max); 
System.out.println(maxI + "," + maxJ); 

Answer 8

//C++ code 
#include<iostream> 
#include<vector> 
#include<algorithm> 
using namespace std; 
vector<int> b; 
vector<int> c; 
int Func(int a[][10],int n) 
{ 
    int max; 
    max=a[0][0]; 
    for(int i=0;i<n;i++) 
    { 
      for(int j=0;j<n;j++) 
      { 
        if(a[i][j]>max) 
        { 
            max=a[i][j]; 
            b.push_back(i); 
            c.push_back(j); 
            } 

        } 
        } 
        b.push_back(0); 
        c.push_back(0); 
        return max; 
        } 
    void display(int a[][10],int n) 
    { 
     for(int i=0;i<n;i++) 
    { 
      for(int j=0;j<n;j++) 
      { 
        cout<<a[i][j]<<"\t"; 
        } 
        cout<<endl; 
        } 
        } 

int main() 
{ 
    int a[10][10],n; 
    cin>>n; 
    for(int i=0;i<n;i++) 
    { 
      for(int j=0;j<n;j++) 
      { 
        cin>>a[i][j]; 
        } 
        } 
        cout<<endl; 
        display(a,n); 
        cout<<endl; 
        cout<<Func(a,n)<<" is the greatest "<<endl; 
        if(b.size()==1&&c.size()==1) 
        { 
              cout<<"Location is (1,1)"<<endl; 
              } 
              else 
              { 
               b.erase(b.end() - 1); 
               c.erase(c.end() - 1); 
        cout<<"Location is "<<"("<<b.back()+1<<","<<c.back()+1<<")"<<endl; 
        } 
        return 0; 
        } 

Answer 9

你只是增加指數i和j在一起，然後將它打印到屏幕。由於您正在運行整個循環，因此它將等於2 * arr.length-2。當您遇到新的最大值時，您需要做的是存儲i和j的值。

例如：

int[][] arr = {{4, 44, 5, 7, 63, 1}, {7, 88, 31, 95, 9, 6}, {88, 99, 6, 5, 77, 4}}; 

int max = arr[0][0]; //dunno why you made it double when you're dealing with integers 
int max_row=0; 
int max_column=0; 
for (int i = 0; i < arr.length; i++) { 
    for (int j = 0; j < arr.length; j++) { 
     if (arr[i][j] > max) { 
      max = arr[i][j]; 
      max_row=i; 
      max_column=j; 

     } 
    } 
System.out.println("The max is: "+max+" at index ["+max_row+"]["+max_column+"]");