本文介紹了遞歸地對數組中的整數求和的處理方法,對大家解決問題具有一定的參考價值,需要的朋友們下面隨著小編來一起學習吧!
問題描述
我有一個程序,我正在嘗試為類創建一個使用遞歸返回數組中所有整數之和的程序.到目前為止,這是我的程序:
I have a program that I'm trying to make for class that returns the sum of all the integers in an array using recursion. Here is my program thus far:
public class SumOfArray {
private int[] a;
private int n;
private int result;
public int sumOfArray(int[] a) {
this.a = a;
n = a.length;
if (n == 0) // base case
result = 0;
else
result = a[n] + sumOfArray(a[n-1]);
return result;
} // End SumOfArray method
} // End SumOfArray Class
但是我相信我遇到了三個相關的錯誤,但我無法弄清楚為什么它會找到一種 null:
But I'm getting three error which are all related, I believe, but I can't figure out why it is finding a type of null:
SumOfArray.java:25: sumOfArray(int[]) in SumOfArray cannot be applied to (int)
result = a[n] + sumOfArray(a[n-1]);
^
SumOfArray.java:25: operator + cannot be applied to int,sumOfArray
result = a[n] + sumOfArray(a[n-1]);
^
SumOfArray.java:25: incompatible types
found : <nulltype>
required: int
result = a[n] + sumOfArray(a[n-1]);
^
3 errors
推薦答案
解決方案比看起來簡單,試試這個(假設一個非零長度的數組):
The solution is simpler than it looks, try this (assuming an array with non-zero length):
public int sumOfArray(int[] a, int n) {
if (n == 0)
return a[n];
else
return a[n] + sumOfArray(a, n-1);
}
這樣稱呼它:
int[] a = { 1, 2, 3, 4, 5 };
int sum = sumOfArray(a, a.length-1);
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