java - 使用 for 循环组合 1 个数组的元素以创建一个新数组

Question

大家下午好，

我被分配了一个编程问题，我需要采用具有不同数量元素的单个数组，并配置一个新数组。

新数组将由前两个元素的和组成，后跟接下来的两个元素的和，依此类推。

如果原始数组有奇数个元素，则最后一个值将保持不变，而求和值在前面。

public static int[] collapse(int[] old) {
    int[] nNew = new int[old.length];        
    if(old.length%2 == 0){
        nNew = new int[old.length/2];    
        for(int i =0; i<nNew.length; i+=2) {                        
            int temp = old[i] + old[i+1];
            nNew[i] = temp;            
        }                   
    }
     //i tried to for loop to increment by 1 here, but it didn't work out.
     //I believe this is where my problem is.

    if(old.length%2 != 0) {
        nNew = new int[(old.length/2) +1];        
        for(int i =0; i<nNew.length-1; i++) {
            //I can't get an Odd number of array elements to increment by 2
            int temp = old[i] + old[i+1];
            nNew[i] = temp;
            nNew[nNew.length-1] = old[old.length-1];
        }                
    }    
    return nNew; 
}

调用示例包括:

collapse({7, 2, 8, 9, 4, 13, 7, 1, 9, 10}) 
expected: {9, 17, 17, 8, 19}
my result: {9, 0, 17, 0, 17}

collapse({1, 2, 3, 4, 5}) 
expected: {3, 7, 5} 
my result: {3, 5, 5}

Answer 1

不需要单独的循环。只需在同一循环内检查即可:

public static int[] collapse(int[] old) {
    int[] nNew = new int[(old.length + 1) / 2];
    for (int i = 0; i < old.length; i += 2) {
        nNew[i / 2] = old[i];
        if (i + 1 < old.length) {
            nNew[i / 2] += old[i + 1];
        }
    }
    return nNew;
}