数组中元素计数的CodingBat帮助

Question

public boolean haveThree(int[] nums) {
    int count=0;
    for (int i = 0; i < nums.length - 1; i++) {
        if (nums[i] == 3 && nums[i + 1] == 3)
            return false;
        if (nums[i] == 3)
            count++;
    }
    if (nums[nums.length-1] == 3)
        count++;
    return(count==3);
}

以下是测试结果：

Expected Run        
have Three({3, 1, 3, 1, 3}) → true  true OK     
have Three({3, 1, 3, 3}) → false    false   OK      
have Three({3, 4, 3, 3, 4}) → false false   OK      
have Three({1,  3, 1, 3, 1, 2}) → false false   OK      
have Three({1, 3, 1, 3, 1, 3}) → true   true    OK      
have Three({1, 3, 3, 1, 3}) → false false   OK      
have Three({1, 3, 1, 3, 1, 3, 4, 3}) → false    false   OK      
have Three({3, 4, 3, 4, 3, 4, 4}) → true    true    OK       
have Three({3, 3, 3}) → false   false   OK      
have Three({1, 3}) → false  false   OK      
have Three({3}) → false false   OK      
have Three({1}) → false false   OK      
other tests X       

Answer 1

这样做太复杂了，只需计算3的出现次数并执行检查：

int count=0;
for(int i=0; i < nums.length; i++){
    if(nums[i]==3)
        count++;
}
return (count == 3);

或者，您可以让事情变得更快一些，因为如果您处于n-1位置，并且您到目前为止只计算了一个或一个3，您知道这可能会失败，因此：

for(int i=0; count <= 3 && (count == 3 || i < nums.length-2+count); i++){
    if(nums[i]==3)
        count++;
}
return count == 3;

当遇到三个以上3时，这也停止了。但我建议不要这样做，因为它会使代码更加不可读。

int count=0;
for(int i=0; i < nums.length-1; i++){
    if(nums[i] == 3 && nums[i+1] == 3) {
        return false;
    }
}
for(int i=0; i < nums.length; i++){
    if(nums[i]==3)
        count++;
}
return (count == 3);

int count=0;
for(int i=0; i < nums.length; i++){
    if(nums[i] == 3) {
        count++;
        if(i+1 < nums.length && nums[i+1] == 3) {
            return false;
        }
    }
}
return (count == 3);

public boolean haveThree (int[] nums) {
    if(nums == null) {
        return false;
    }
    int count=0;
    for(int i=0; i < nums.length && count <= 3; i++){
        if(nums[i] == 3) {
            count++;
            if(i+1 < nums.length && nums[i+1] == 3) {
                return false;
            }
        }
    }
    return (count == 3);
}

if (nums[nums.length-1] == 3)
    count++;

如果(nums[-1]==3)now-1不作为索引存在，则将导致对的查询，因此IndexoutofboundsException。我在CodeBat上测试了这一点，实际上，这似乎是问题所在。

解决这个问题的一种方法是添加一个“零长度”检查：

public boolean haveThree(int[] nums) {
    if(nums.length <= 0) //hint, although a length is never negative, one better uses "strong conditions"
        return false;
    int count=0;
    for (int i = 0; i < nums.length - 1; i++) {
        if (nums[i] == 3 && nums[i + 1] == 3)
            return false;
        if (nums[i] == 3)
            count++;
    }
    if (nums[nums.length-1] == 3)
        count++;
    return(count==3);
}

但我建议使用上面所述的版本，因为总的来说，这些方法可以减少错误。