吹拉弹唱


  • Home
  • Archive
  • Categories
  • Tags
  • Books
  •  

© 2022 Kleon

Theme Typography by Makito

Proudly published with Hexo

Code - Number

Posted at 2022-05-23Updated at 2022-05-23 interview  interview algorithm 

  • 7. Reverse Integer
  • 8. String to Integer (atoi)
  • 65. Valid Number
  • 191. Number of 1 Bits
  • 190. Reverse Bits
  • 2119. A Number After a Double Reversal

# 7. Reverse Integer

leetcode

We could solve by reversing it as a string. But the problem here is we have to check overflow in a 32-bit environment.

We can either use (0xffffffff - tail)/10 to prevent overflow, or use the reversion to check overflow.

1
2
3
4
5
6
7
8
9
10
11
12
13
14
func reverse(x int) int {
    var result int32 = 0 
    for x != 0 {
        var tail int32 = int32(x % 10)
        var newResult int32 = result * 10 + tail
        if (newResult - tail) / 10 != result {
            return 0
        }
        result = newResult
        x /= 10
    }
    
    return int(result)
}

# 8. String to Integer (atoi)

leetcode

1
2
3
4
5
6
7
8
9
10
11
12
13
func myAtoi(s string) int {
    r := regexp.MustCompile(`^[ ]*[-+]?[\d]+`)
    s = string(r.Find([]byte(s)))
    s = strings.TrimSpace(s)
    // fmt.Printf("%v", s)
    num, _ := strconv.Atoi(s)
    if num > int(math.Pow(2, 31) - 1) {
        return int(math.Pow(2, 31) - 1)
    } else if num < int(-math.Pow(2, 31)) {
        return int(-math.Pow(2, 31))
    }
    return num
}

# 65. Valid Number

leetcode

Regexp

1
2
3
4
func isNumber(s string) bool {
    r := regexp.MustCompile(`^[+-]?((\d+\.?\d*)|(\d*\.?\d+))([eE][+-]?\d+)?$`)
    return r.MatchString(s)
}

condition traverse

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
func isNumber(s string) bool {
    s = strings.TrimSpace(s)
    pointSeen := false
    eSeen := false
    numberSeen := false
    
    for i := 0; i < len(s); i++ {
        if s[i] <= '9' && s[i] >= '0' {
            numberSeen = true
        } else if s[i] == '.' {
            if eSeen || pointSeen {
                return false
            }
            pointSeen = true
        } else if s[i] == 'e' || s[i] == 'E' {
            if eSeen || !numberSeen {
                return false
            }
            eSeen = true
            numberSeen = false
        } else if s[i] == '-' || s[i] == '+' {
            if i > 0 && s[i-1] != 'e' {
                return false
            }
        } else {
            return false
        }
    }
    return numberSeen
}

# 191. Number of 1 Bits

leetcode

1
2
3
4
5
6
7
8
func hammingWeight(num uint32) int {
   var count uint32 = 0
   for num !=0{
      count += num & 1
      num >>= 1
   }
    return int(count)
}

# 190. Reverse Bits

leetcode

1
2
3
4
5
6
7
8
9
10
func reverseBits(num uint32) uint32 {
    var rev uint32 = 0
    for i := 0; i < 31; i++ {
        rev += num & 1
        num >>= 1
        rev <<= 1
    }
    rev += num
    return rev
}

# 2119. A Number After a Double Reversal

1
2
3
4
5
6
func isSameAfterReversals(num int) bool {
    if num == 0 {
        return true
    }
    return num % 10 != 0
}

Share 

 Previous post: System - YouTube Next post: System Design 

© 2022 Kleon

Theme Typography by Makito

Proudly published with Hexo