【CodingBat】tenRun问题的两种解法

问题描述：tenRun问题

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}.

examples:
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]

代码1：

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

代码2：更简洁

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;
}