法1.采用递归,利用逻辑与的短路特性
public class Solution { int Sum_Solution(int n) { int ans = n; ans && (ans += Sum_Solution(n - 1)); return ans; } }
法2. 调用Java库函数
public class Solution { public int Sum_Solution(int n) { int sum = (int) (Math.pow(n,2) + n); return sum>>1; } }