Java递归求和1+2+3+...+n
public class Sum { public static int count(int n) { if (n > 1) { return count(n - 1) + n; } else { return n; } } public static void main(String[] args) { System.out.println(Sum.count(100)); }}
扩展学习
输入一个数:
4
10
代码:
public class Demo1 {public static void main(String[] args) {System.out.print("请输入一个数:");Scanner sc = new Scanner(System.in);int n = sc.nextInt();Sum(n);System.out.println(Sum(n));}private static int Sum(int n) {if (n == 0) {return 0;}return Sum(n - 1) + n;}}
思路:
计算前n个数的总和等于第n-1个数+n;
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