剑指No1

Posted 2020-12-20 teyues

在一个二维数组中（每个一维数组的长度相同），每一行都按照从左到右递增的顺序排序，每一列都按照从上到下递增的顺序排序。请完成一个函数，输入这样的一个二维数组和一个整数，判断数组中是否含有该整数。
public static void  main(String [] args){
    int value = 563; 
    int times = 0;
    int n = 10;
    int arry[][] = new int[n][n];
    //随机生成满足条件的二维数组
    for(int i = 0;i<arry.length;i++){
      for(int j=0;j<arry[i].length;j++){
          if(j==0) {
              if(i == 0) {
                  arry[i][j] = (int) (Math.random() * 100) + 1;
              }else{
                  int rowgap = (int)(Math.random()*100)+1;
                  arry[i][j] = arry[i-1][j]+rowgap;
              }
          }else{
              if(i ==0){
                  int columgap = (int)(Math.random()*100)+1;
                  arry[i][j] = arry[i][j-1]+columgap;
              }else{
                  int rowcolgap = (int)(Math.random()*100)+1;
                  if(arry[i-1][j]<arry[i][j-1]){
                      arry[i][j] = arry[i][j-1]+rowcolgap;
                  }else{
                      arry[i][j] = arry[i-1][j]+rowcolgap;
                  }
              }
          }
      }
    }
    //遍历输出数组 检验数组是否构造正确
    for( int i = 0 ;i<arry.length;i++){
        for(int j = 0;j<arry[i].length;j++){
            if(j<n-1){
                System.out.print(arry[i][j]+"  ");
            }else{
                System.out.println(arry[i][j]);
            }
        }
    }
    //遍历查找二维数组中是否存在该值
    int row = arry.length -1 ;
    int colum = 0;
    for (int k = row;k>=0; k --){
        for(int b = colum ;b<arry.length; b ++){
            if(value == (arry[row][colum])){
                times =times +1;
                System.out.print("Exist the value of "+value+"查找次数为"+times);
            }else if(value>(arry[row][colum])){
                times = times + 1;
                colum =colum + 1;
            }else{
                times = times + 1;
                row = row -1;
                break;
            }
        }
    }
    System.out.println("查找次数为"+times+"Not exist the value of "+value);
}