[Swift]LeetCode371. 兩整數之和 | Sum of Two Integers
阿新 • • 發佈:2018-10-11
style ive The class lee forward operator 使用 test case
Calculate the sum of two integers a and b, but you are not allowed to use the operator +
and -
.
Example:
Given a = 1 and b = 2, return 3.
Credits:
Special thanks to @fujiaozhu for adding this problem and creating all test cases.
不使用運算符 +
和 -
???????,計算兩整數 ???????a
、b
???????之和。
示例 1:
輸入: a = 1, b = 2 輸出: 3
示例 2:
輸入: a = -2, b = 3 輸出: 1
8ms
1 class Solution { 2 func getSum(_ a: Int, _ b: Int) -> Int { 3 //按位取異或 4 var res:Int = a^b 5 //判斷是否需要進位 6 var forward = (a&b) << 1 7 if forward != 0 8 { 9 //如有進位,則將二進制數左移一位,進行遞歸10 return getSum(res, forward) 11 } 12 return res 13 } 14 }
8ms
1 class Solution { 2 func getSum(_ a: Int, _ b: Int) -> Int { 3 if a == 0 { 4 return b 5 } 6 7 if b == 0 { 8 return a 9 }10 11 var a = a 12 var b = b 13 while b != 0 { 14 let carry = a & b 15 a = a ^ b 16 b = carry << 1 17 } 18 return a 19 } 20 }
8ms
1 class Solution { 2 func getSum(_ a: Int, _ b: Int) -> Int { 3 if a&b==0 { 4 return a|b 5 } 6 return getSum(a^b, (a&b)<<1) 7 } 8 }
12ms
1 class Solution { 2 func getSum(_ a: Int, _ b: Int) -> Int { 3 var carry = (a & b) << 1 4 var result = a ^ b 5 while carry != 0 { 6 let carryTemp = carry 7 carry = (result & carryTemp) << 1 8 result = result ^ carryTemp 9 } 10 return result 11 } 12 }
16ms
1 class Solution { 2 func getSum(_ a: Int, _ b: Int) -> Int { 3 4 if b == 0 { 5 return a 6 } else { 7 return getSum(a ^ b, (a & b) << 1) 8 } 9 } 10 }
[Swift]LeetCode371. 兩整數之和 | Sum of Two Integers