[LeetCode By Go 23]258. Add Digits

题目

Given a non-negative integer num, repeatedly add all its digits until the result has only one digit.

For example:

Given num = 38, the process is like: 3 + 8 = 11, 1 + 1 = 2. Since 2 has only one digit, return it.

Follow up:
Could you do it without any loop/recursion in O(1) runtime?

解题思路

循环累加每一位,时间复杂度O(logN)
不需要循环,递归的暂时没想到,/(ㄒoㄒ)/~~

代码

addDigits.go

package _258_Add_Digits

func AddDigits(num int) int {
    for ;num >= 10; {
        var sum int
        for ;num >= 10; {
            sum += num % 10
            num = num / 10
        }
        num += sum
    }

    return num
}

测试

addDigits_test.go

package _258_Add_Digits

import "testing"

func TestAddDigits(t *testing.T) {
    var tests = []struct{
        num int
        output int
    }{
        {38, 2},
        {10, 1},
    }

    for _, test := range tests {
        ret := AddDigits(test.num)

        if ret == test.output {
            t.Logf("pass")
        } else {
            t.Errorf("fail, want %+v, get %+v", test.output, ret)
        }
    }
}
    原文作者:miltonsun
    原文地址: https://www.jianshu.com/p/4084a723abbe
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