我用java刷 leetcode 剑指 Offer 64. 求1+2+…+n

Posted 2021-06-02 深林无鹿

这里有leetcode题集分类整理！！！

我的憨憨AC：

class Solution {
    public int sumNums(int n) {
        int time = n / 2;
        int ztime = n % 2;
        int res = (1+n) * time + ztime * ((n + 1) / 2);
        return res;
    }
}

官方：
改造递归为 && 逻辑短路

class Solution {
    public int sumNums(int n) {
        boolean flag = n > 0 && (n += sumNums(n - 1)) > 0;
        return n;
    }
}

作者：LeetCode-Solution
链接：https://leetcode-cn.com/problems/qiu-12n-lcof/solution/qiu-12n-by-leetcode-solution/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。

快速乘：

class Solution {
    public int sumNums(int n) {
        int ans = 0, A = n, B = n + 1;
        boolean flag;

        flag = ((B & 1) > 0) && (ans += A) > 0;
        A <<= 1;
        B >>= 1;

        flag = ((B & 1) > 0) && (ans += A) > 0;
        A <<= 1;
        B >>= 1;

        flag = ((B & 1) > 0) && (ans += A) > 0;
        A <<= 1;
        B >>= 1;

        flag = ((B & 1) > 0) && (ans += A) > 0;
        A <<= 1;
        B >>= 1;

        flag = ((B & 1) > 0) && (ans += A) > 0;
        A <<= 1;
        B >>= 1;

        flag = ((B & 1) > 0) && (ans += A) > 0;
        A <<= 1;
        B >>= 1;

        flag = ((B & 1) > 0) && (ans += A) > 0;
        A <<= 1;
        B >>= 1;

        flag = ((B & 1) > 0) && (ans += A) > 0;
        A <<= 1;
        B >>= 1;

        flag = ((B & 1) > 0) && (ans += A) > 0;
        A <<= 1;
        B >>= 1;

        flag = ((B & 1) > 0) && (ans += A) > 0;
        A <<= 1;
        B >>= 1;

        flag = ((B & 1) > 0) && (ans += A) > 0;
        A <<= 1;
        B >>= 1;

        flag = ((B & 1) > 0) && (ans += A) > 0;
        A <<= 1;
        B >>= 1;

        flag = ((B & 1) > 0) && (ans += A) > 0;
        A <<= 1;
        B >>= 1;

        flag = ((B & 1) > 0) && (ans += A) > 0;
        A <<= 1;
        B >>= 1;

        return ans >> 1;
    }
}

作者：LeetCode-Solution
链接：https://leetcode-cn.com/problems/qiu-12n-lcof/solution/qiu-12n-by-leetcode-solution/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。

数学 + 快速乘

class Solution {
    public int sumNums(int n) {
        return quickMul(n, n + 1) >> 1;
    }
    int quickMul(int a, int b){
        int res = 0;
        boolean flag;
        flag = b != 0 &&
              (flag = (b & 1) == 1 && (res += a + quickMul(a<<=1, b>>=1)) > 0 ||
              (res += quickMul(a<<=1, b>>=1)) > 0);
        return res;
    }
}