Java > Array-2 > tenRun (CodingBat Solution)

Problem:

For each multiple of 10 in the given array, change all the values following it to be that multiple of 10, until encountering another multiple of 10. So {2, 10, 3, 4, 20, 5} yields {2, 10, 10, 10, 20, 20}.

tenRun({2, 10, 3, 4, 20, 5}) → {2, 10, 10, 10, 20, 20}
tenRun({10, 1, 20, 2}) → {10, 10, 20, 20}
tenRun({10, 1, 9, 20}) → {10, 10, 10, 20}


Solution:

public int[] tenRun(int[] nums) {
  boolean ten = false;
  int tmp = 0;
  
  for (int i = 0; i < nums.length; i++) {
    if (nums[i] % 10 == 0) {
      tmp = nums[i];
      ten = true;
    }
    else if (nums[i] % 10 != 0 && ten) {
      nums[i] = tmp;
    } 
  }
  return nums;
}


5 comments :

  1. public int[] tenRun(int[] nums) {
    int temp = 0;
    for (int i=0; i<nums.length-1; i++)
    {
    if (nums[i]%10==0 && nums[i+1]%10!=0) nums[i+1] = nums[i];
    }
    return nums;
    }

    ReplyDelete
  2. public int[] tenRun(int[] nums) {
    int val = 1;
    boolean wasTen = false;

    for(int i = 0; i < nums.length; i++) {
    if(nums[i] % 10 == 0 && nums[i] != val) {
    val = nums[i];
    wasTen = true;
    }
    if(wasTen) nums[i] = val;
    }

    return nums;

    }

    ReplyDelete
  3. public int[] tenRun(int[] nums) {
    for (int i=0; i<nums.length-1; i++) {
    if (nums[i]%10==0 && nums[i+1]%10!=0) {
    nums[i+1] = nums[i];
    }
    }
    return nums;
    }

    ReplyDelete
  4. public int[] tenRun(int[] nums) {
    for (int i=0; i<nums.length-1; i++) {
    if (nums[i]%10==0 && nums[i+1]%10!=0) {
    nums[i+1] = nums[i];
    }
    }
    return nums;
    }

    ReplyDelete
  5. public int[] tenRun(int[] nums) {

    for(int i=0 ; i<nums.length-1 ;i++){
    if(nums[i]%10==0){

    int temp = nums[i];

    while(i<nums.length-1 && nums[i+1]%10!=0){
    nums[i+1]=temp;
    i++;
    }
    }
    }
    return nums;
    }

    ReplyDelete

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